help-36

[soft zealot]

• VUR0 ◇ DUSSEKAR •

[mighty grotto]

?

[tender pollen]

yep

distance from J to O?

[mighty grotto]

4.5

[tender pollen]

not quite

how'd you get that?

[tender pollen]

i wouldn't approximate

[mighty grotto]

after i get 4.472135 what do i do

[tender pollen]

well, that's still approximating

can you find the exact value?

[mighty grotto]

uhh

maybe its the numbers

[tender pollen]

try evaluating the squares first

what do you get?

[mighty grotto]

(4-2)^2 = 4

(2+6)^2 = 64

[tender pollen]

add them now

[mighty grotto]

radical 68 is 8.2462112512

[tender pollen]

what's the exact value?

[mighty grotto]

that is

[tender pollen]

well actually

i think sqrt(68) is fine

what about the other distance from O to E?

[mighty grotto]

hollup maybe the a's and b's are frying me

is there an x and y for this

ill search

[tender pollen]

?

[mighty grotto]

idk where the place the numbers so looking for and x and y version might be easier

ohh

[tender pollen]

yeah

[mighty grotto]

ok hollon ima get scrap paper

[tender pollen]

!noai though lol

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[mighty grotto]

sorry

[tender pollen]

it's ok

it was correct so i'm pretty sure it's fine

[mighty grotto]

(-2-4)^2+(4-6) =

radical 40

about 6.3

but my teacher said always keep it in radical

[tender pollen]

yes

keep it in radical for exact answers

[mighty grotto]

gotchu

[tender pollen]

J to E?

[mighty grotto]

(6-(-2))^2 = 64 then (0-4)^2= 16

thats radical 80

[tender pollen]

wait

did you make a mistake on one of them?

[tender pollen]

i think you messed up this

[mighty grotto]

oh i missed a square

[tender pollen]

oh nvm

it's correct

[mighty grotto]

okur

[tender pollen]

but there's a calculation error

somewhere

[mighty grotto]

do i do the same for the 3rd side

oh

what error so i can fix

[tender pollen]

yeah i think it's this one?

check (4-6)

[mighty grotto]

i meant 4-6 square so thats -2 * -2 which is 4

[tender pollen]

the content in 4-6

(||it's not 4||)

[mighty grotto]

ou

do i need to switch the 6 and 4 around

[tender pollen]

no

the coordinate isn't 4

it's ||0||

[mighty grotto]

Huh

[tender pollen]

wait what

how'd you get #2?

(6-(-2))^2 + (0-4)^2

everything else seems correct

[mighty grotto]

from one side to antoher side

ok ok lemme redo i think i see it

ah shucks

i got radical 80 this time

[tender pollen]

nice

[mighty grotto]

which means i prolly gotta switch the values

[tender pollen]

so which sides are equal?

[mighty grotto]

JE and JO are equal

[tender pollen]

yes, so...

__/\__JOE is...

[mighty grotto]

isoceles?

[tender pollen]

yes

[tender pollen]

also, not to be the grammar police, but it's spelled "iso__s__celes"

[mighty grotto]

O thanks i didnt catch that

[tender pollen]

ok

so we know that JE = JO

so what does J lie on?

[mighty grotto]

i dont understand the question

[tender pollen]

i'm talking about the second one now

[mighty grotto]

oh the bisector

uhh it align

s

with 2,2

which is Y

[tender pollen]

yep

suppose X lies on the perpendicular bisector of OE

what can we say about X?

[mighty grotto]

that X got bisected aswell

[tender pollen]

so?

what does it mean about XO and XE?

[mighty grotto]

i genuinely have no idea

that theyre bisectors?

[tender pollen]

ok

as it turns out,

it's a theorem in geometry that

the perpendicular bisector of AB contains all points C

such that CA = CB

and contains ONLY those points

therefore, C lies on the perpendicular bisector of AB if and only if CA = CB

and we know that JE = JO already

so what can we say about YJ?

(hint: ||two points form a line||)

[mighty grotto]

YJ is a right angle?

[tender pollen]

???

a line can't be a right angle

lol

[tender pollen]

what i'm trying to say is

Y and J lie on the perpendicular bisector of OE

and the perpendicular bisector is a line

so YJ must form that line

since two points form a line

[mighty grotto]

ohhhhhhh

i see i see

[tender pollen]

so that's the proof of #2

if you're done type ".close"

[mighty grotto]

so overall, for#2 id say that J and Y line up as a bisector and go trough OE?

[tender pollen]

yes

J and Y lie on the perp. bisector

and the perp. bisector is a line

so J and Y must form that line

[tender pollen]

(proof: prove that JO = JE and YO = YE)

[mighty grotto]

SORRY I WAS COOKING SOMETHING

[mighty grotto]

THANK YOUU

[tender pollen]

if you're done type ".close"

[mighty grotto]

si si

.close

[final saddle]

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[hushed basin]

can anyone help me with the top 1

[hushed basin]

Just a little hint on what i should do

[deep condor]

,rccw

[deep condor]

are you able to do any of it, or do you need help on starting?

[hushed basin]

start of it

i completely forgot anything related to that

Yea there is smth i am diff missing

[ornate sky]

You know there is a result/formula for displacement in nth second in straight line motion

you may recall deriving it using second equation of motion

[trail mango]

very mid handwriting

[mental patrol]

Blud is still stuck at handwriting

[ornate sky]

if that is mid then mine in unfathomable

[mental patrol]

S(4) - S(3) = 26
S(9) - S(8) = 56

$S(t)=ut+\frac{1}{2}at^2$

[soft zealot]

◇♡《pasta》♤♧

[hushed basin]

Va=Vo+(n-1_2)a

[ornate sky]

Snth=u+a/2 (2n-1)

[hushed basin]

sry i am panicking the assignment is due 40 mins

[hushed basin]

what is u

[trail mango]

it's ok

[ornate sky]

Initial velocity

a is the constant acc.

[hushed basin]

s is disp

[ornate sky]

n is the nth second

[hushed basin]

yea my teach never taught us that lmfao

[ornate sky]

just apply it two time. You get two equation and two variables

[hushed basin]

S/t =Va , Va=Vo+(n-1_2)a

[ornate sky]

then use second equation of motion. like @mental patrol showed

[hushed basin]

r the only thing i know and idk how to use em

[hushed basin]

elaborate pls?

[ornate sky]

I meant input the two set conditions in the result.

[hushed basin]

i am dumb

i understood absolutely nothing

[hushed basin]

ok so

[ornate sky]

uh first set of condition is Snth=26m and n=4. Put these values in the result and you will get a equation in a and u.

[hushed basin]

lets say 26 x 4

= ...

[ornate sky]

and do the same for second set of conditions. You get another such equation.

[hushed basin]

what is snth again

[ornate sky]

Displacement in nth second

yeah so two equations with u and a as variable. Find them. That's what is asked

[hushed basin]

26= 3.5a

right

?

[ornate sky]

you forgot u

[hushed basin]

isn't it equal 0

start of motion?

[ornate sky]

No. Read the ques properly. obviously at some point the particle started from rest but at t=0 or the time when we started observing the motion, it had attained a velocity which we call its initial velocity

[hushed basin]

ic

[ornate sky]

??

did you solve the equations

I did and got the answers as well. so you can too

[hushed basin]

oh my pain

i got it

[hushed basin]

thank u

[hushed basin]

I did

On g whzt is this
what about multiple Sec th

[unreal vessel]

this is for 1st quection answer u can do it like this . do u need other answers also

[hushed basin]

ic

i did basically the same thing and omitted

[hushed basin]

i did in the 2nd question

8th to 10th

so i did

8+9+10/3

×0.5

ah shit - 0.5 mb

right

8.5

Letsgo i got the hang of it

thanks yall

.close

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[barren loom]

The problem asks to find length of arc AB if the angle PRB is equal to 46 and S and R are points of tangency

Can anyone help me solve this problem?

[primal sapphire]

okay, whats that?

[barren loom]

Circumference inscribed in a circumference

[cosmic warren]

For starters, what arc measures can you determine with angle PRB?

[barren loom]

Arc pb

And angle pab

[barren loom]

Arc pb is twice 46

So 92

[cosmic warren]

yup!

[barren loom]

Wait

If I draw a tangent line such that it touches B

Is that line parallel to PA?

[cosmic warren]

I'm not sure if it necessarily should or shouldn't be but why do you want to draw that tangent?

[barren loom]

Because that tangent will split the arc PA into 2 equal parts

Waitt

So PB = AB??

[cosmic warren]

Yup

[barren loom]

Is there any other way to reach that conclusion

[cosmic warren]

I got that in a different manner though haha

Yeah there is

[barren loom]

Interesting

Mind showing how?

[cosmic warren]

You label point Q as such

Then QRB = (RP +PB)/2 = minor arc RS

Also since S is tangent you have minor arc RS = angle RSP = (RP + AB)/2

Oops typo

[barren loom]

How is the minor arc RS the same as arc RB?

[cosmic warren]

Sorry I mean

Because RS is subtended by tangent line QR and RS

And RB is subtended by tangent line RQ and line RB which is the same as RS

Key is that R is tangent to both circles

[barren loom]

Could you keep explain in dms if you don’t mind? I might go offline suddenly

[cosmic warren]

Ok

[barren loom]

Power in my house is going out lately

[final saddle]

@barren loom Has your question been resolved?

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[sullen echo]

plz help with this

[gaunt raft]

!status

[final saddle]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@sullen echo Has your question been resolved?

[lilac bison]

Ever heard of the king's property?

[final saddle]

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[spice vine]

Guys im so lost on how i can do the rest of the lines to finish up the art

[spice vine]

can you guys give me ideass on what kind of functions i can use

heres some of the requirements

[royal gust]

You cannot do odd functions, since odd functions have to pass through the origin

[stone wagon]

1/x doesn't pass through the origin

[royal gust]

You have to include asymptotes? That's a wild thing to include

[royal gust]

Fair! Mb

[stone wagon]

[spice vine]

Yeah man

im cooked

I can do restrictions tho

[royal gust]

You can move the pikachu so that there's lines through the origin

That will make it easier to do some odd functions

I guess it's not a big deal to include asymptotes as long as you're allowed to restrict them

I didn't think of that haha

[spice vine]

i see

[severe hawk]

what's wrong with two metric fucktons of polynomials?

also, the "list of requirements" surely is missing stuff. You cannot be asked for two "horizontal asymptotes" if there's literally zero horizontal lines in the drawing

[spice vine]

we had to pick our own image

[royal gust]

You can sneak a few in there! You'll just have to restrict them to be very small

[spice vine]

yeahh

[final saddle]

@spice vine Has your question been resolved?

[final saddle]

@spice vine Has your question been resolved?

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[raven loom]

Can you help me in understanding point c? I don't know what I have to do in this question tbh

[raven loom]

I solved a and b, but c is like not clear on what to do

[abstract bramble]

for what values of x is f(x) <= 1

[raven loom]

$x >= 4$

[soft zealot]

JPuXIUim6x

[raven loom]

Oh yeah show me latex special language syntax

[abstract bramble]

is it >=?

well you should get it right first

[raven loom]

I mean, appearantly I am wrong, with solving it

[abstract bramble]

its asking <= 1

[raven loom]

The answer sheet says it's the other direction

LIke $x =< 4$

[soft zealot]

JPuXIUim6x

[abstract bramble]

yes because... the f(x) or y value is <= 1 there

im not sure how you can flip the sign

refer to the graph

when x <= 4 then f(x) <= 1

[raven loom]

It's asking me to solve inequality

[abstract bramble]

uh huh

[raven loom]

Not to explain it on a graph

So I need some proof

Or validation

[abstract bramble]

no you dont

the graph is the proof

you're just writing an observation based on it

solving inequalities is putting it in the form of
"x sign value"

thats all you need to be concerned about for now

[raven loom]

So for $f(x) =< 1, x =< 4$?

[soft zealot]

JPuXIUim6x

[abstract bramble]

i would reverse the order of that statement

[raven loom]

??

[abstract bramble]

For $x \leq 4, f(x) \leq 1$

[raven loom]

is this domain correct?

Because I solved it, but appearantly my answer sheet says it's not correct

[soft zealot]

tten ʚɞ

[abstract bramble]

what did you put for domain

[raven loom]

$4 \leq x$

[soft zealot]

JPuXIUim6x

[abstract bramble]

huh...

thats just your original answer

[raven loom]

That's what I got

[abstract bramble]

yes and i told you

refer to the graph

[raven loom]

Is it wrong?

[abstract bramble]

is your answer right

yes its wrong

look at the graph

[raven loom]

Yes

[abstract bramble]

for x > 4

is the value of the function less than or equal to 1

[raven loom]

fx will be 1 and smaller

[abstract bramble]

no it isnt

its literally

greater than 1

[raven loom]

For f(x) smaller or equal to 1, x will be equal to 4 or less

[abstract bramble]

Okay, so why did you write this

[raven loom]

Because I solved the inequalty wrong, I flipped the sign when was multiplying

[abstract bramble]

this is correct but as i said it's best to reverse the order (causality)

usually you would say

for x blah blah blah, f(x) blah blah blah

because f(x) depends on x

[raven loom]

Makes sense

But for some reason

Those are the answers

Why did they write -2 I have no idea

[abstract bramble]

oh right

because past -2 it's greater than 1

[raven loom]

Asymptote?

[abstract bramble]

well yeah just check the graph

[raven loom]

Ok

Thank you

A little confused still

But cleared up

[forest sage]

see the graph

[abstract bramble]

draw a line y = 1

look at what x values f(x) is below the line

[raven loom]

3, 2, 1, 0

And 4

Intersection

And till x = -2

Since this moment is asymptote, it will go until infinity

.close

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[copper roost]

hello, so im stuck here

[copper roost]

im not sure if i should factorize x, or use the pq formula

for x1,2

[finite tree]

wdym?

[tiny gorge]

how about the quadratic formula

[finite tree]

you can't really factorize, just use the

yeah

[copper roost]

wait why cant i tho

x(x-6+9c+18)

[finite tree]

you can but you'd end up with quadratic formula, there's no nice factorization

[tiny gorge]

18 doesn't have an x attached to it, that factorization is wrong

[copper roost]

oh so how can i like memorize the rule to that? every number must have an x in order to be factorized?

[finite tree]

in this case it's more or so you can take x "common" or take it outside of an expression only if it has an x attached to it yeah

also just memorize the quadratic formula if you haven't gone over it, you can find the solution to any quadratic that way

[sturdy nacelle]

Theres no rule to be memorised. The simple logic is that your factors must multiply together to get the original expression

[vital crag]

,tex .quadratic formula

[soft zealot]

riemann

[sturdy nacelle]

In this case it simply doesnt

[vital crag]

this is probably the one you should memorize

[sturdy nacelle]

This case

[copper roost]

ohhh

okay got it

[copper roost]

i memorized that

its just

here, my b is 9c+18 right

[tiny gorge]

b is the coefficient of the x term

[finite tree]

your quadratic is
x^2-6x+9c+18
correct?

[tiny gorge]

your 18 does not have an x attached

[copper roost]

yed

[copper roost]

yeah thats my b then

no

wtf

omg

[tiny gorge]

your 18 is part of the c in the quadratic formula

[copper roost]

and my 9c is also the c in the formula right

[tiny gorge]

confusingly, you also have c in your expression, so you probably want to rename something

[finite tree]

yep

[tiny gorge]

because writing "c = 9c + 18" is going to be a recipe for misery

[copper roost]

hold up ill put it into the formula i know

[copper roost]

here

[tiny gorge]

ok so now how will you proceed?

[copper roost]

so it says determine the number of points with a horizontal tangent so

maybe like a case differentiation

thingy

for D=0 and stuff?

[tiny gorge]

if D is the thing inside the square root then yes

[copper roost]

ohhh

so if D=0 then we have 1 simple point with x=3

otherwise if D>0 then we have 2 points and D<0 then no zeropoint?

[tiny gorge]

yep correct

[copper roost]

but do i have to look at how c is specifically in the cases? like c>0 etc

?

[tiny gorge]

yes, translate your three cases (D = 0, D > 0, D < 0) into corresponding statements involving c

[copper roost]

shit ok

[lilac bison]

good handwriting

[trail mango]

don’t steal my shtick

[lilac bison]

what

[trail mango]

also

medium handwriting

[lilac bison]

it was a satirical comment

[trail mango]

unsolicited handwriting feedback is a slaylaism. you cannot adopt it

[lilac bison]

Law, you have great handwriting, don't let her comments get to you, but yeah you can improve

[lilac bison]

i dont see a trademark sign anywhere

[trail mango]

ts is not great

it’s medium

[little python]

good handwriting :3

[lilac bison]

🫂

[trail mango]

i’m so done

[lilac bison]

by the power of democracy you lose here

[hybrid heath]

Wonderul handwriting

[lilac bison]

my brethren

istg, ill get banned someday..

[copper roost]

people are so mean

[finite tree]

you can say they are average

[copper roost]

my handwriting sucks ik but

[trail mango]

good one aorts

[finite tree]

🥁 tss

[clever oxide]

its the internet, people can be mean with no consequence lol

[finite tree]

thank u thank u, i will see myself out

[lilac bison]

why is this becoming a full blown discussion

😭

[copper roost]

fuck my handwriting i dont need it the results are what matter

[trail mango]

my opinion counts more since i am a handwriting authority

[lilac bison]

i mean, its important its comprehensible no?

[clever oxide]

but if people can't read it then they can't get a result tho

[lilac bison]

uh okay?

[copper roost]

yes but people can read it

[lilac bison]

https://klipy.com/gifs/congratulations-congratulations-gif-1

[clever oxide]

i mean im not sure if its 78 or 18, and is that a c or a u ?

[lilac bison]

what was the original question again? 💀

[copper roost]

okay

[lilac bison]

is this 18?

or 78?

[copper roost]

so im kind of stuck

[copper roost]

18

so if d = 0 then c =-1

but wouldnt that mean we have x 1 and x 2 = 3

???

[severe canyon]

Well, they must be indeed equal

[copper roost]

because we have 3 + , - square root of 0

[severe canyon]

Yeah this should be something very well known from quadratics

[copper roost]

so if d=0 then x1,2 = 3? not just x1=3?

[tiny gorge]

because the quadratic factors as (x-3)(x-3) in that case

so there's information being conveyed by referring to x1,x2

[inland widget]

hi guys

[copper roost]

so?

[tiny gorge]

but for your purpose, all you care about is that this is the case where there is one solution

[copper roost]

hmm

[severe canyon]

I mean, how would you write the solutions to x² - 4x + 4 = 0??

[copper roost]

yeah x1,2 = …

but if i had to sketch the determinant then how do i draw a double zeropoint in a tangent line

damn

[tiny gorge]

since you're doing calculus, i'll mention as a side note that when a polynomial has a "double root" (like x1,x2 = 3) then 3 is a root of both the polynomial and its derivative

there are contexts where that is of interest

[copper roost]

oh

hm okay

[copper roost]

okay so ive done it like this

basically when d>0 and c<-1 then we have 2 different zeropoints

when c>-1 then none

but im a bit confused

[tiny gorge]

seems ok, what's the confusion?

i would word it slightly differently: instead of "when d > 0 and c < -1" i would say "if c < -1, then d > 0, so there are two solutions"

the way you said it suggests that d and c are independent of each other, but they are not

[copper roost]

so is there like a trick for finding out the c<.. part

because

i had to like actually think

and put values into c

bcs i got like 45 minutes for the whole exam

[tiny gorge]

well you can just solve the inequality like this:
D > 0
<==> 9 - (9c + 18) > 0
<==> -9c - 9 > 0
<==> 9c < -9
<==> c < -1

[copper roost]

or i could just like think about when a positive number would come out from the stuff inside

[tiny gorge]

yea you can do it that way too

[copper roost]

ok nice

[tiny gorge]

or as you said, just plug in a number for c that is greater than -1 and see what you get

[copper roost]

also my teacher does some stuff with

]-1 ; infinity

or some stuff

i forgot that notation

[tiny gorge]

yea that's just another way of saying c > -1

[copper roost]

oh okay

ty tho rly helped me

[tiny gorge]

yw

[final saddle]

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[cyan kayak]

So do you know the formula for a weighted average?

[warped phoenix]

Uh no

[frozen sparrow]

i need help with something, im new in this server so i dont know if im in the right channel

its math

[cyan kayak]

!help

[frozen sparrow]

and about areas

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[cyan kayak]

So a formula for a weighted average is $\frac{ax + by}{a + b}$ where $x$ and $y$ are the values that you're taking the average of, and $a$ and $b$ are the weights.

[soft zealot]

OmnipotentEntity

[warped phoenix]

I see

[cyan kayak]

so what we have is $\frac{a \cdot 0.5 + b \cdot 0.25}{a + b} = 0.4$

[soft zealot]

OmnipotentEntity

[cyan kayak]

now, $a$ and $b$ can be general weights, but we can simplify this by recognizing that we let $a$ and $b$ have the meaning of the volume used.

[soft zealot]

OmnipotentEntity

[cyan kayak]

So @warped phoenix can you make a second formula which relates only a to b in a different way?

using the idea above?

[warped phoenix]

Hmm lemme see

[abstract bramble]

Read the last line of the original problem

[warped phoenix]

Ahh i still cant grasp it

Ive never encountered a question like this before

[cyan kayak]

that's OK!

that's the feeling you get when you're learning

in the last sentence of the original problem, how much volume total are you shooting for?

[warped phoenix]

10 liters

[cyan kayak]

ok, so if a is the volume of the 50% concentration acid, and b is the volume of 25% concentration acid, then 10 liters represents what combination of a and b?

[warped phoenix]

a + b = 10

[cyan kayak]

exactly

So now you have two equations and two unknowns

[warped phoenix]

Yess i can do this by my self now i think

x = 6, y = 4?

I use x and y for simplicity

a = 6, y = 4

Uh do i close

.close

[final saddle]

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[cyan kayak]

This is correct.

[final saddle]

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[winter lava]

$\lim_{n\to\infty}\sum_{k=0}^{n}f(n)^k = \frac{1}{1-\lim_{n\to\infty} f(n)}$

[soft zealot]

Axe

[winter lava]

this is allowed?

[ebon agate]

Weird notation

ig if the limit exists then yes

[vital crag]

did you try doing an intermediate step with partial sums formula

[winter lava]

oh

[ebon agate]

Can you explain. I thought they just used the GP formula

[winter lava]

that could be a way to justify it

[strange fractal]

They did, but the question is - can you use it. It would not be unreasonable to suspect that the convergence needs to be uniform in order for it to be allowed.

The TLD;DR is that you need to think about it well before doing so.

[ebon agate]

Right

Makes sense

[winter lava]

like this, right?

thanks

.solved

[final saddle]

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[winter lava]

.reopen

[final saddle]

✅ Original question: #help-36 message

[winter lava]

my mistake

the exponent on the second line

ok it still works as f(n)^(n+1) doesn't give any indeterminate form

.close

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[rain sentinel]

Okay so I still need help clarifying the argument in the second and third quadrants

[rain sentinel]

So for these examples the argument = theta

So fir the second quadrant theta=pi-alpha
And third quadrant
Theta=pi+alpha

[opal plinth]

Are you talking about the argument of a complex number, as in arg(z)?

[rain sentinel]

Yes argument in modulus argument form

[opal plinth]

It's always the angle with the positive x axis

[rain sentinel]

So

[opal plinth]

It's mod 2pi and can be considered within (-pi, pi] but the starting point of the angle stays the same

[rain sentinel]

Right

So my diagram is correct?

Nvm

It's incorrect

So is the argument the arrow?

[stray hull]

Yes

[rain sentinel]

Ok ty

[final saddle]

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[final saddle]

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[hardy jackal]

is this correct

[abstract bramble]

appears so

but i would use a more obvious colour than dark blue

could barely see it

[hardy jackal]

okok

thank u

.close

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[rain sentinel]

can someone explain what cos6theta + jsin6theta goes

or is it equal to 1

[vital crag]

did you try plugging in 3pi / 4 for theta

[rain sentinel]

yeah

so it becomes 4096(cos6pi + jsin6pi)

[vital crag]

do you know the values of cos and sin of (6pi) ?

[rain sentinel]

yh

hmm

[rain sentinel]

but where does j go?

[vital crag]

0 times anything equals ?

[rain sentinel]

ahhh

thanks

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[distant inlet]

How do i do this? Explain in ez words im braindead

[final saddle]

@distant inlet Has your question been resolved?

[distant inlet]

gang

[distant inlet]

.close

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[hybrid slate]

you want the 3rd quartile, or 75th percentile, or top 25%, so the probability in invnorm is 0.75 (75th percentile). so with the other parameters you should have invnorm(0.75, 119, 16)

[whole hazel]

Exactly, I did that, but it gave me the same exact answer as Q1, and I do NOT know what went wrong lol

I pasted it to the right, if that was incorrect

[hybrid slate]

well you could compute without it then, you have the mean and standard deviation, so just with the z score for the 75th percentile you can calculate it normally i.e. no shortcut functions

[whole hazel]

So I just have to find the z-score?

[hybrid slate]

yeah, typically in class (or at least in mine) you'd get a table of z scores

so it shouldnt be disallowed knowledge or anything

[whole hazel]

I see. I don't have a table of z-scores or anything. I was just told to use the invnorm function

I didn't even know you could find that from calculating the z-score

[hybrid slate]

i will say im not sure why you're getting 108.21 for both Q1 and Q3 using invnorm

[whole hazel]

I got -7.39 for the z-score

[whole hazel]

I don't know either. I pasted left for Q1, then right for Q3, but it got me the same answer for some reason

[hybrid slate]

well you shouldn't be swapping the tail

[whole hazel]

The tail, the lowercase sigma, standard deviation?

I did Z = 0.75 - 119 / 16

[hybrid slate]

if you look it up in a table you'll find this for the z score

[whole hazel]

Help, what

Yeah, I never knew that

[hybrid slate]

the z score formula won't really help because you're solving for the x-value

you rearrange it to get x = mean + (z * std_deviation)

[whole hazel]

X-value? I thought it was q3

[hybrid slate]

yes we're solving for the x-value at q3

[whole hazel]

Ah

Yeah, I don't seem to have that formula here

[hybrid slate]

I guess they want you to use invnorm only for some reason ¯_(ツ)_/¯

[whole hazel]

Yeah, I don't know

She said that because the written calculations can sometimes be messy and take up time so we use calculators, TI-84s

[whole hazel]

So the answer is 129.78 after doing this calculation

[hybrid slate]

something like that

[whole hazel]

Mkay

Lemme try

[hybrid slate]

you should be getting something similar with invnorm

[whole hazel]

Agh no its not

[hybrid slate]

im not sure whats different between the two

[whole hazel]

What the freak is going on

I don't know, I'm genuinely so lost lol

[hybrid slate]

just try rounding the decimal up or down by a thousandth lol

hundredth*

[whole hazel]

Likeee.. To 130?

[hybrid slate]

no by 0.01

[whole hazel]

It wants two decimal places

That would still be 129.784, no?

So then 129.78

[hybrid slate]

129.79

[whole hazel]

O

It can't be 79 tho, cuz 4 is too small to round it

[hybrid slate]

give it a shot

[whole hazel]

Okay

How the fuck

Sorry

HOW IS THAT RIGHT LMAO

Im SO confused 😭

[hybrid slate]

z-score precision

[whole hazel]

How does the rounding work that way tho

[hybrid slate]

fwiw this is what i was talking about with invnorm

[whole hazel]

I see

Yeah, we don't have that lol

I dunno what is going on with this math anymore

[hybrid slate]

it doesn't, but if you recalculated with the more accurate z-score of 0.6745 you'd get the appropriate decimal which rounds to .79

[whole hazel]

I see, I see

Yeah, I didn't know I was supposed to use a site, I just calculated with what you gave me lol

My bad

[hybrid slate]

its fine im hardly learned in statistics since its not important to my major/career

[whole hazel]

That's fair

[hybrid slate]

i would recommend figuring out why invnorm wasn't working out for you though

[whole hazel]

It's not to mine either, I don't think, it's just required for my major and is good if I switch my major, I guess

I'm trying to do psychology bro

[hybrid slate]

lol

[whole hazel]

Will do, I'll ask my prof!

Thank you for the help :,)

.close

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[plush haven]

help

[plush haven]

why did the top intergrand change from 1 to -1

[lilac bison]

Good handwriting

[final tangle]

upper bound/limit, not integrand

[cyan kayak]

@plush haven when we do u substitutions we must also change the bounds of the integral. So we go from x = 0 to u = 0, but when x = 1, then u = -x^2 = -1

(we also substituted for the lower bound it just happened to not change)

[plush haven]

oh

.close

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[toxic pike]

I need help with a geometry problem which I'm like 90% done

[toxic pike]

Given triangle ABC inscribed in circle (O). T is an arbitary point on (BOC). The line thorugh T perpendicular to OT intersects BC at X. Construct paralellogram ATXY. AY intersects BC at Z. Prove that AX and ZT intersect at a point on the radical axis of (BOC) and (XYZ).

I only need TA^2 = TZ.TF

Also AX intersects (XYZ) at a second point E, TZ intersects (BOC) at a second point F

[gleaming shore]

i had a heart attack seeing the graph

ping helpers after 15 mins

[toxic pike]

ok ty

[final saddle]

@toxic pike Has your question been resolved?

[toxic pike]

<@&286206848099549185>

[final saddle]

@toxic pike Has your question been resolved?

[stark minnow]

me not a geometry pro, is TA a tangent to the leftmost circle ?

if yes, then this is a direct result of the tangent secant theorem

[toxic pike]

Thats what we need to prove rn yes

[final saddle]

@toxic pike Has your question been resolved?

[final saddle]

@toxic pike Has your question been resolved?

[final saddle]

@toxic pike Has your question been resolved?

[final saddle]

@toxic pike Has your question been resolved?

[cedar kraken]

[hushed bluff]

I think you should ask gemini or chatgpt

[toxic pike]

I dont believe in AI’s capability of solving geometry problems

For now

[civic terrace]

That's the worst thing you can ask an AI.

AI sucks in geometry.

[spark swan]

<@&268886789983436800>

[gritty chasm]

may not work, but this looks suspiciously the same as if in a circle ABP is a secant and PT is a tangent for some exterior point P, then $PT^2 = PA\cdot PB$

[soft zealot]

Annie Maqionde

[toxic pike]

Yes

But we need to prove that TA is the tangent

[white hinge]

exteextend TA from point A on the left circle

[final saddle]

@toxic pike Has your question been resolved?

[toxic pike]

It wont cut yes but we need to actually prove it

[tranquil pine]

heart attack 😭

[tulip shard]

is this question a real question?

[leaden moon]

woah woah woah woah

what is the need for you to ping here

-# ||oh yeah he blocked me for no reason kek||

[sharp ruin]

What grade is that.

[vital crag]

<@&268886789983436800> spam

[toxic pike]

Yes

[toxic pike]

I dont really know tbh

[brittle kestrel]

Help

[jagged flare]

if you can get DAB+DCB=TAC then i got a solution

[radiant igloo]

Holy. The fact that I've seen shit worse than this...

[idle sorrel]

Would maybe explode the circles to make their curvature 0. Then it would just be angle chasing I think

[ripe flicker]

sob

[barren pebble]

Hi, do you still need help? If so, on which part?

[fathom jackal]

this question will return on doomsday 🙏

[finite tree]

Have you tried an analytic approach @toxic pike

[terse vine]

😭yo put a spolier or smth

[toxic pike]

💀

[toxic pike]

I have no idea what that is

[drifting current]

It just a bunch of lines bro

I can’t even read this

[toxic pike]

[mortal nexus]

This puts mental health of people on this server in danger bro😭

[sturdy cypress]

discord is a not a safe place in general

[honest widget]

With is that question

[toxic pike]

😶

I guess at your place geometry is different

Probably some of the Descartes Oxy stuffs

[finite tree]

Oh I meant as in analytical geometry, like using coordinate geometry
there's some pretty nice properties like S1-S2=0 being the radical axis of two circles where S1 and S2 stand for the circle equations/power of a point, T=0 being the tangent equation for a point where there's a nice transformation you can do to S to obtain it, and more
it would be very algebra heavy and grindy but possibly worth it if you haven't got a solution yet

[toxic pike]

Oh I understand but I’m not allowed to use that thing on this problem

[frigid gate]

@toxic pike IMO ?

[toxic pike]

Noo

Nowhere near

It's just some radical axis practice problems

[frigid gate]

What the question

[frigid gate]

Vietnamese Math?

[toxic pike]

Ya

[frigid gate]

Đợi tí

Nhìn như đề IMO :vv

[toxic pike]

💀

Never expected that there are so many VNese people here

[frigid gate]

Có giả thiết m câu trc ko

[frigid gate]

Đợi sáng mai có kq

[toxic pike]

This is the whole question

[toxic pike]

😃 ok

[frigid gate]

Dậy t ra rồi

Bây giờ muốn TA^2=TZ.TF

Thì TA.TA=TZ.TF

=> TA/TF = TZ/TA

Mà ATXY là hình bình hành(Given)

=> AZ//TX

=> góc TAZ = góc ATX ( 2 góc so le trong)

Xét đg tròn BOC có

TX là tiếp tuyến tại F và TF là dây cung

=> góc ATX = góc TBF

Vì TX là tiếp tuyến BOC

Và AZ//TX

Do t/c hbh ATXY

[toxic pike]

I wonder how TX is tangent to (BOC)

[frigid gate]

Đg tròn BOC có tâm O

Mà T nằm trên đg tròn BOC

=> OT là bán kinha

[toxic pike]

What

T is an arbitary point on (BOC)

[frigid gate]

Có OT là dây cung

Của đg tròn BÕC

Đk là TX vg OT

=> TX là tiếp tuyến

[toxic pike]

You're saying TX is tangent of (BOC)?

[red tree]

where’s the original problem statement screenshot

[toxic pike]

it's in vietnamese

[low otter]

what on earth is this thing

[toxic pike]

geometry 😄

[lofty nexus]

try asking on aops

[sweet socket]

geometry jumpscare

[severe hawk]

!original

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[azure pasture]

this is your time of dying

[leaden moon]

send it, ill translate to yall

[toxic pike]

I already translated it to English

[leaden moon]

.

original problem

[toxic pike]

[mortal bay]

hmmmm

can u draw a better one?

ý tôi là bạn có thể vẽ hình bé hơn được không, cái đường tròn BOC to quá :/

[lapis flax]

holy geom

[lapis flax]

what's the original version

[frigid gate]

Vietnamese geometry welcome

[simple topaz]

how do i solve that

[vagrant folio]

Funny enough, my lecture a few weeks ago was on that!

[mental patrol]

math oly shyt?

[toxic pike]

Given triangle ABC inscribed in circle (O). T is an arbitary point on (BOC). The line thorugh T perpendicular to OT intersects BC at X. Construct paralellogram ATXY. AY intersects BC at Z. Prove that AX and ZT intersect at a point on the radical axis of (BOC) and (XYZ).

[mental patrol]

circle centered at O?

and T is a point where exactly? on the arc?

[toxic pike]

y

e

[gaunt raft]

Wth is this

[muted glade]

I mean im vietnamese grade 10 and I'm pretty familiar with these

My weakest area is 2D geo so I usually dont don't do these but usually you do one part then similar prove for the others

That's the case with most cluttered problem but sometimes you do have to use every info

[chrome bramble]

Guys can y'all help me to solve this problem

[tranquil pine]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help ⁠ for instructions).

[muted glade]

if you mean just 3^40 + 3^40 + 3^40 part i think it's just 3^41

[muted glade]

i mean my idea is to prove TA is tangent to (AZF) but i think you have to get more extra lines

[toxic pike]

Yes

[muted glade]

since there isn't anything know by AF

so there's no way it's solvable with the current shape

[muted glade]

also how does proving TA^2 = TZ*TF solve the problem

[toxic pike]

AX intersects TZ at D

DX intersects (XYZ) at E, TD intersects (BOC) at F

Therefore we need to prove that TXFE is a cyclic quadrilateral

Sorry I had to go grab some stuffs

[toxic pike]

This is equivalent to EFT = EXT

EXT = EAZ

So we need to prove that EFT=EAZ

Or AEZF is a cyclic quadrilateral

or AFZ=ZEX

ZEX=ZAX+AZE=ZAX+YXE=ZAX+XAT=ZAT

So we need to prove AFZ=ZAT

or TA^2 = TF*TZ

And then I'm stuck ther

e

[barren jasper]

Ur cooked

[proper dagger]

I don't think that is a very helpful comment to make.

[tranquil jackal]

<@&268886789983436800> Commenting in many help channels, unhelpfully

[tulip coyote]

(already got dealt with )

[civic terrace]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[muted glade]

K I asked the problem to my teacher and he solved it (credit to NguyenLePhuoc on Facebook)

Let AZ cut (O) in N, we have AN // TX so TA = TN

After that AFNT is cyclic which lead to AEZF is cyclic

[polar saffron]

same lol tho

[tranquil jackal]

@toxic pike Have you made any progress?

[tranquil pine]

Ur every where@tranquil jackal

[tranquil jackal]

I look around

[timber leaf]

well first step i would do is showing AZ // TX

Since $Z\in AY$

[soft zealot]

Minhh

[timber leaf]

But, TX is perpendicular to OT (as given), so AZ also perpendicular to OT

now we will need to show that OTXZ is in a circle

Consider the circle centre O where the ABC inside, we see that OB=OC= radius

Then thecircle where OBC (O not centre) is symmetrical across the perpendicular bisector BC

But TX perpendiclar to OT and AZ perpendicular to OT that you can prove this quadrilateral inside that circle

[civic terrace]

Wait we're solving this question right?

[timber leaf]

ye boss

[civic terrace]

Let me state it clear first.

I didn't construct this diagram.

So we are supposed to prove that P lies on the radical axis right?

[neon lotus]

yes

[tranquil jackal]

How hasn't this channel timed out?

[inland kettle]

still no somehow damn

[tranquil pine]

@toxic pike do you still need help

[toxic pike]

Oh It's done

[tranquil pine]

then you should close this

[toxic pike]

.close

[final saddle]

Channel closed

Closed by @toxic pike

Use .reopen if this was a mistake.

[toxic pike]

ok

ty

[lilac bison]

you're welcome

[tranquil jackal]

This has been closed for 7 hrs how isn't it available yet

[inland kettle]

idek

[molten terrace]

<@&286206848099549185>

[tranquil jackal]

If you need help open a channel

[vital crag]

#❓how-to-get-help

[molten terrace]

Aight mb

[tranquil jackal]

Can we get someone to close this because I see no reason for it to still be open

[tranquil jackal]

Even after it's been closed

[strange sparrow]

This is by design

A long channel will stay open for a little while after it's been closed

[tranquil jackal]

Where does it say that 🤔

[strange sparrow]

It's not documented if that's what you mean

[tranquil jackal]

🙁

My bad then G

[tranquil pine]

Mhm

https://tenor.com/view/vae-gif-15357506855217945854

[lime flame]

[blissful edge]

<@&286206848099549185> i need help with this equation

[leaden moon]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tranquil jackal]

It's not occupied

[leaden moon]

oh nvm

[timber leaf]

Type the bot too fast

[tranquil jackal]

This channel is in an evil limbo

[leaden moon]

but next time, go to an empty channel like #help-14

[tranquil jackal]

Not quite open, not quite closed

[blissful edge]

Ok good to know

[tranquil jackal]

https://giphy.com/gifs/gd-limbo-geometry-dash-iQyUSq11icLE93hBrY

[tranquil pine]

dw there was one which was here for four days after it was closed

[odd ruin]

focus

[tranquil pine]

this one will go faster

[pliant elk]

this has been going on and on for 7 days

will this ever end

[whole halo]

.reopen

[final saddle]

✅ Original question: #help-36 message

[whole halo]

.close reset timer

[final saddle]

Channel closed

Closed by @whole halo

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pine lance]

does anyone have the time to list which formulas i need to use to solve these?

[pine lance]

so far i have these

and these

[wild stag]

From left to right
1- The x^n integral
1- And the e^x Intgeral
2- algabric factorization of quadratic
Second picture
1- x^n Intgeral
2- multiplie the x to the root
The down picture
1- x^n
2- e^ax Intgeral

[pine lance]

by 1 and 2 do you mean a and b?

[wild stag]

Yes

[pine lance]

great thank you so much

[wild stag]

Any time mate

[pine lance]

.closed

.close

[final saddle]

Channel closed

Closed by @pine lance

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[red tree]

is it possible to split a circlr into three areas of equal area with 2 NON intersecting chords

[timber leaf]

No bro

[red tree]

how come

just curious

[timber leaf]

wait actually you can

by making 2 parallelels chords but must be on the opposite sides of the centre

and each of these segments must make 1/3 area, which is compulsory

and to show this you cant just use ruler or normal calculator but need to use Area of a segment of a circle's formula

[sturdy cypress]

they don't have to be parrallel either, it's just harder to do

[timber leaf]

well you can but i think parallel would be easier

[final saddle]

@red tree Has your question been resolved?

[red tree]

i see