help-26

[exotic dome]

Now you know the whole lower part

[orchid hatch]

yea

54

[exotic dome]

And the upper part is 1/4 of the whole triangle (similar triangles)

So the lower part is 3/4 of the whole triangle

So multiply the lower part by 4/3 to get the whole triangle

[orchid hatch]

yea

s0

72

[exotic dome]

https://tenor.com/view/magic-confetti-awesome-gif-11884906

[orchid hatch]

3 more

Three squares with sides of lengths two, four and six units, respectively, are arranged
side-by-side, as shown so that one side of each square lies on line AB and a segment connects the
bottom left corner of the smallest square to the upper right corner of the largest square. What is the
area of the shaded quadrilateral?

[exotic dome]

I never do the math though so that’s on you

[orchid hatch]

i cant wait to sleep

[orchid hatch]

yea

[exotic dome]

Ok

[orchid hatch]

im stuck on this

[exotic dome]

Use Thales to find the question mark lengths

[orchid hatch]

?

uhh

ok

idk tht

[exotic dome]

Let x be the small one and y the big one

[orchid hatch]

i think its 18 the entire thing

wiat nvm

[exotic dome]

x/6 =2/(2+4+6)

Right ?

[orchid hatch]

ye

[exotic dome]

x = 1

[orchid hatch]

mhmm

then

[exotic dome]

y /6 = (2+4)/(2+4+6)

[orchid hatch]

3

[exotic dome]

y = 3

[orchid hatch]

area is 8

[exotic dome]

So now you can compute the area right ?

[orchid hatch]

yea

((1+3)*4)/2

[exotic dome]

I never do the math

That’s on you

[orchid hatch]

ye lol

In triangle ABC, AC = 21, BC = 28, and ∠ACB = 90◦. The bisector of ∠ACB meets AB at D.
Find the length BD

hmm

c = 35

ratio is /49

[exotic dome]

Need to go

Brb

[orchid hatch]

huh

[limber elbow]

you need help with this ?

[orchid hatch]

AD is 15

[orchid hatch]

i need someone to verify

AD is 15

BD is 20

[limber elbow]

okay wait a sec

[orchid hatch]

Using the same diagram as in Problem 12, find CD

[orchid hatch]

CD is uh

[limber elbow]

do you have the ratio or should i use trigonometry?

like about the cd

[orchid hatch]

is AD and CD correct

[limber elbow]

what angle it makes or any information about it

[orchid hatch]

i mean is AD =15 and BD = 20

i think CD is also 20

[orchid hatch]

for?

[limber elbow]

okay soo untill you tell me the relation of angle c with the line cd it's not possible to determine it

like if it's bisected then it's fine we can find it

[orchid hatch]

yea

[limber elbow]

soo angle c is bisected ?

[orchid hatch]

In triangle ABC, AC = 21, BC = 28, and ∠ACB = 90◦. The bisector of ∠ACB meets AB at D.
Find the length BD.

[limber elbow]

that makes it easy

[orchid hatch]

ye

[orchid hatch]

is AD 15?

and is BD 20?

[limber elbow]

yeah then you are right my friend

[orchid hatch]

then waht is CD?

[limber elbow]

TBH you got me rolling how is this even solvable

[orchid hatch]

lol

[orchid hatch]

AD = 15; BD = 20; CD =?

[exotic dome]

Are you still on this one V

[orchid hatch]

?

[orchid hatch]

im on this

i got AD = 15; BD = 20; I need to find CD

[limber elbow]

just done

wait a sec

[orchid hatch]

i think its 20 ngl

[orchid hatch]

?

[orchid hatch]

bro messaged at 7:11

[limber elbow]

it's 12*sqrt(2)

[orchid hatch]

https://tenor.com/view/conveniencestores-7eleven-seveneleven-gif-6040957

[orchid hatch]

root(384) i got

[orchid hatch]

how

wait nvm

i figured mistake

[limber elbow]

use the angle bisector theorem

[orchid hatch]

In the diagram, triangles BQA, BQP, PQD, and BPC are all 30◦-60◦-90◦ triangles. If
AD = 12, then find the area of rectangle ABCD.

[orchid hatch]

ja

[limber elbow]

i am sorry but what does 30◦-60◦-90◦ triangles means ?

[orchid hatch]

one angle is 30 another is 60 and last is 90

[limber elbow]

oh kk

[limber elbow]

i am sorry but are you german ?

[orchid hatch]

kindof lol

i think it is 144root(3)

[orchid hatch]

have you gotten anywhere

[limber elbow]

actually i am close

i used rote push

[orchid hatch]

ok

[limber elbow]

naming with variables

[orchid hatch]

ok

[limber elbow]

hey you can use co-ordinates i guess it's 144*sqrt(3)

[orchid hatch]

ye

In rectangle ABCD, P is a point on side AB, and Q is the intersection of BD and CP.
If the area of triangle PBQ is 8, and the area of triangle CDQ is 18, then find the area of rectangle
ABCD

2nd last one

i got 60

[limber elbow]

yeah

you're right

[orchid hatch]

ok

In the diagram, ABCD is a square with side length 13, DP = BQ = 5, and
AP = CQ = 12. Find PQ (((((i got sqrt(178)))))

hmm

do u have any clue

i kinda guessed too

[limber elbow]

nah i am solving RN

[orchid hatch]

ok

[limber elbow]

might take a bit i am using co-ordinates and stuff

[orchid hatch]

ye

[orchid hatch]

did u get anywhere

[limber elbow]

17*sqrt(2)

just let me

cehck ones

yeah

17*sqrt(2)

@orchid hatch

[orchid hatch]

ok

[limber elbow]

want me to send the solution ?

[orchid hatch]

sure

[limber elbow]

kk

[orchid hatch]

i gtg

im not too sure its 17root(2) but i think it is that

[limber elbow]

place the square with coordinates (D at origin):
A = (0, 13), B = (13, 13), C = (13, 0), D = (0, 0).

Point P is the intersection of the circles centered at A and D with radii AP = 12 and DP = 5.
Solving:

(x – 0)² + (y – 13)² = 12² ( using the cricle equation )

x² + y² = 5²
gives the left intersection:
P = (–60/13, 25/13).

Point Q is the intersection of the circles centered at B and C with radii BQ = 5 and CQ = 12.
Solving:

(x – 13)² + (y – 13)² = 5²

(x – 13)² + y² = 12²
and taking the point to the right of the square gives:
Q = (229/13, 144/13).

Now compute the vector PQ = (289/13, 119/13).

Then,
PQ² = (289/13)² + (119/13)²
= (289² + 119²) / 169
= 97682 / 169
= 578
= 2 × 17²

PQ = 17√2.

there you go took me a while to get it from image to text correctly cause my handwriting is aah soo good haha !

[orchid hatch]

oh lol

ok ig

ty

[limber elbow]

welcome mate

[orchid hatch]

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[plush seal]

3. Compare the V(i) that you calculated for times 7, 13, 21, and 25s. with the Vavg for the intervals that included those times. in which cases are the V(i) and Vavg nearly the same? quite different? Explain why.

Since the instantanious velocity is the tangent of a position time curve, it can be calculated by the infinitesimal change in position with respect to time. Since we are only given a finite number of points, calculating the tangent at a p(7) is approximately the same as finding the secant of the two points closest to p(7) (e.g: p(6), p(8)). Therefore, since we are only given a finite set of points, Vavg and V(i) are approximately the same.

[plush seal]

i have

quite literally

no idea if my analysis was right

[patent owl]

SPH3UI.. godspeed soldier

[plush seal]

my teacher does NOT deserve this level of analysis but we do what we do 😔

acc this is for an in class lab i didnt even submit 💀

[patent owl]

i think the idea is, you want to find V_avg for the specific phase, and compare it with V_i in each case.

[plush seal]

here’s the graph

you can barely read it tho

that graph is for

vavg

the table is for position

[patent owl]

no that's perfectly fine

[patent owl]

so for t=7, get the secant line between 6 and 8, but then get the secant line over all of phase 1 (this is the v_avg i think they want)

and then you should be comparing 5 different times, with each interval, to see when it's close and when it's far

[plush seal]

pretty sure they would all be the exact same

since im just calculating for vi

and in this case

vi = vavg

so vavg=vavg

yeah i cross referenced

all of these are almost the exact same

[patent owl]

with the Vavg for the intervals that included those times
i interpreted this as being the velocity over each phase, since that actually has the potential to be different

so for phase 1, it's the average velocity between 0.0 and 8.0 seconds

i mean i could be wrong about this wording. but like you said, vi = vavg, and i doubt they would give you a question that easily

[plush seal]

i mean we had to calculate the average velocity of every point on the position time graph

then graph the average velocity

which is where that graph came in

iroc is basically just

the closest two points to a given curve gives you the tangent of the curve

which is basically

we are making the secant so much smaller

that it basically becomes the tangent

which is essentially what im doing

if we expand the iroc's tangent

we get a secant

and since vavg is a secant from all the points on the position graph

if we calculated for vi on a point

it would be the same as vavg

since we do the exact same calculation

because we cant find the actual derivitive

anyways that just how i interpreted it as

i might be wrong plus its hella late

[patent owl]

hmm, im actually gonna double back and say that your thinking makes more sense here, because they also included 13s which is on the cusp of each phase

[plush seal]

im just going to act like im right and move on from this shitty course

alr lol thank you

[patent owl]

in that case, your analysis is pretty much correct. i might word your second sentence like this, though:

Since we are only given a finite number of points, our closest approximation for the tangent at p(i) is finding the secant....
and you could also add to the ending sentence:
since we are only given a finite set of points, V_avg = V_i within the closest possible approximation

but i think you get the point across fine without those changes

[patent owl]

also my 4U teacher was quite literally a skeleton and he retired early, immediately after teaching our class. so yeah 💀

(but i honestly think 4U was easier than 3U, because the bs rules start to make a bit more sense)

[plush seal]

to be fair

i have like a 50 average in this course because

im taking 4U next sem

so this wont count to my uni marks at all

so who gives a shit

[plush seal]

yeah i kinda wrote it all on pencil so its hella hard to write lol

[patent owl]

probably the best attitude to have about the course. hope u learn something cool in the meanwhile at least

[plush seal]

i mean i love physics but this god damn shit show of a teacher man

not to flame her like

[patent owl]

maybe you can keep these words in the back of your head for later then

[plush seal]

she prolly a good teacher for some people

but like

we spent 2 hours on a single lesson

which took me

15 minutes of self study

then she assigned 15 homework problems EVERY DAY 💀

[patent owl]

"closest approximation" and "within the given degree of precision" i think describe these kinds of things rly well

i do NOT miss the amount of mandatory SPH homework

[plush seal]

what you end with

[plush seal]

so like

i havent done any of it

theres 15 days of homework

ive gotten 0/5s on all of em

[patent owl]

its been a while and i forgot what my transcript looked like. i think i got 91-93 for both courses bc i did all the work (it was covid i had nothing else to do)

[plush seal]

nicee

im at a solid 60

okay tbf

thats not my fault

[patent owl]

ya if u can get the work in, you should

[plush seal]

on my quiz

i wrote x1 = (fraction1) x2 = (fraction2)

and the equation was in time

so i got a ZERO on that ENTIRE PART

so wouldve been a 100

got like

a 75 i think

or an 80 on that

[patent owl]

and this is why i do not like physics

in the first-year uni course for physics at my school, they gave us the standard "6-step method to solve a problem" thing

they wanted full explanations for each step for each problem, or else you lose like half your marks

and the multiple choice was 15 questions out of 5 points. no half points.

[plush seal]

yeah cause all of the marks are like only process work bullshit

physics pisses me off but like as a subject

its better than math

[patent owl]

im kind of whining now. but i really don't understand why they care so much, if the method and answer are correct

[plush seal]

bro my ENTIRE answer was right but i just forgot to do t1= and t2=

[patent owl]

well this is a math server so

[plush seal]

holy shit this teacher

[plush seal]

true true

theres not many physics servers tho

[patent owl]

no sympathy from me unfortunately

[plush seal]

i dont blame you lol

[patent owl]

yeah apparently the big one is dead?

[plush seal]

yeah

alr ima go

[patent owl]

sounds good, good luck ^-^

[plush seal]

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

I'm wondering if I went about proving injectiveness right

in the back of the book the solution said it follows from r^-1 existing

I didn't use that but is mine also right?

[outer portal]

think of 2 * 5 = 0 (mod 10)

Oh nvm completely misread

[outer portal]

No, “it can’t be 0” isn’t how you show that x - y = 0 (mod n)

think of 2 * 5 = 0 (mod 10)

[grand solar]

thats true

[outer portal]

You can just use that r^-1 exists on both sides

also (mod n) is written once per equation instead of each side

[grand solar]

oh ok

are we multiplying both side by r^-1?

[outer portal]

yes

[grand solar]

hmm

is the operation were dealing with addition or multiplication for Z_n

[outer portal]

It’s just the same operation as between s and r

[grand solar]

does r inverse exist because U(n) is a group?

or does it exists because Z_n is a group?

r is just some integer mod n so even if it's inverse didn't exist in U(n) I would still be able to use it's inverse from Z_n?

[topaz sinew]

@grand solar Has your question been resolved?

[grand solar]

im stuck on proving homomorphism

[valid marsh]

Z_n is made into a group with addition, but we use the fact that we can multiply integers here

[valid marsh]

this follows by distributivity of integer multiplication

[grand solar]

wait so

oh ok

so sr is multiplication but were in addition

[valid marsh]

it exists because we assume that r is in U(n), and by the definition of U(n), r must have a (multiplicative) inverse mod n

[grand solar]

so alpha(xy) is alpha(x + y)

[valid marsh]

No, notice that the domain of alpha is Z_n

So you need to show that alpha is a homomorphism from (Z_n, +) to itself

and notice that the group operation is + in both the domain and the codomain

[grand solar]

oh ok that makes sense

[grand solar]

what if we defined r to be in the set of U(n) excluding r's inverse

would that mean we would not be able to assume the inverse exist

I was thinking since were dealing with Z_n and r is just an integer mod n we could just still assume it inverse existed

[valid marsh]

do you know the definition of U(n)?

[grand solar]

U(n) is numbers relatively prime to n under multiplication mod n

[valid marsh]

2 does not have an inverse mod 10, so the assumption that r is in U(n) is necessary

[grand solar]

well I never knew if multiplication mod n is a part of U(n)

[valid marsh]

multiplication mod n is a group operation on U(n)

addition isn’t (on U(n))

[valid marsh]

this involves the Euclidean Algorithm or Bezout’s Identity or something, if we use your definition

because it is a theorem that every r which is coprime to n must have an inverse mod n

[grand solar]

I think im a bit confused still here

why wouldn't 8 be the inverse of 2 mod 10?

[valid marsh]

oh, I mean 2 does not have a multiplicative inverse

[valid marsh]

for “additive inverse” I’d say “negative”

[quasi depot]

8*2 is not 1 mod 10

[grand solar]

but in Z_n 2 has an inverse mod 10 right

I think im confused on why we cant use that inverse

[valid marsh]

usually “inverse… mod n” (for some n) would mean “multiplicative inverse”, since the additive inverse is extremely easy to find, so we say “negative” for that

[valid marsh]

use it where?

[grand solar]

so we have sr for r in U(n), and say I want to show alpha is injective
then xr = yr, and I want to multiply

-- oh I think I see, I was confused on why we coudn't use it here, but thats because were multiplying here and 2 inverse in Z_n is not the same as 2 inverse in U(n)? There's a lot going on so my brain is getting kind of jumbled thinking of different operations

[quasi depot]

Its the same difference between -2 and 1/2

[valid marsh]

yeah so you can consider calling them “inverse” for “multiplicative inverse” and “negative” for “additive inverse”

[grand solar]

ok ill keep that in mind, thanks for the help guys

.solved

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[grand solar]

so I want to show that G is isomorphic to H?

and both of their group operations is addition

yeah this question has me kind of lost

[loud oasis]

take G and H as groups under addition. then prove they are isomorphic

[grand solar]

So I need to find a function that relates the two groups, ok ill think some more about it

though kind of confused on how I would relate a real number with a matrix but hmm

ill try and see if I can get something

maybe det or something

oo I think det might do it?

a^2 - 2b^2

and thats (a - sqrt(2))(a + sqrt(2)b)

ill see if I can use this

[loud oasis]

i would say the notation is suggestive here

[topaz sinew]

@grand solar Has your question been resolved?

[clear python]

And then check if the isomorphism preserves multiplication

[grand solar]

did i show Onto correctly?

[clear python]

Yup looks good

Now you have to check the next part

[grand solar]

ok thanks i’ll try

[grand solar]

when they say prove G and H are closed under multiplication they mean for me to show that two elements in G or H should produce another element in G or H under multiplication right

[clear python]

Yes

[grand solar]

ok

[grand solar]

thank you for help, onto my last question

.solved

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[stark crater]

how to solve the integral of (l+a+x)^-2

[stark crater]

I dont know about u du method there was I role I think that can be followed to solve such thing

[maiden plank]

$\int_0^L (l+a+x)^{-2} \dd x$?

[stark crater]

not 1 L

[maiden plank]

oh

[rough furnace]

what's L

[stark crater]

yes

[maiden plank]

are l and a some constants?

[stark crater]

basically part of line charge

[stark crater]

yes u can assume so

and integeral limits from 0 to L

if that would help

[thorny flame]

Céline

[stark crater]

yes that is it

[rough furnace]

$\int \frac{1}{(c+x)^{2}} \dd x$

[thorny flame]

Alexis_Fx

[rough furnace]

Do this look familiar to you?

[stark crater]

yea same form as our problem

but it is in denominator instead

[rough furnace]

What's the differential of -1/(x+c)

[stark crater]

wait.. that was arc tan smth right?

arctan(x/a)/ sqrt(a) maybe

wait I think am wrong

[rough furnace]

just answer this

[stark crater]

ok -1

nah wait

-ln (x)

[rough furnace]

no....

[marsh charm]

alright what is the antiderivative of x

[stark crater]

we subtract -1 from the x+c that would give us (x+c)^-2

[rough furnace]

yes differential of -1/(x+c) is (x+c)^-2

So antiderivative of (x+c)^-2 is -1/(x+c)

Does this look familiar to you?

[stark crater]

alr.. it is same as our problem

so it should be

-1/(l+a+x)

[rough furnace]

yessss

That's not the way we suppose to solve integral but it works just fine in this case lol

[stark crater]

there is no rule for this case?

I mean

[rough furnace]

Yes we do, I don't remember which one tho

[stark crater]

when we get antiderivative of (x+c)^n we should do that and such ?

I see..

that works too

[rough furnace]

$\int (x+c)^n=\frac{1}{n+1} (x+c)^{n+1}$

[thorny flame]

Alexis_Fx

[rough furnace]

iirc

[stark crater]

oh wait

so I was watching this video

and when they solved this integration he added 1 , which makes sense.. Then mulitplied by -1

I belive it is the differential of l+a-x maybe

[rough furnace]

you mean the yellow circle?

[stark crater]

no ignore the yello circle

[rough furnace]

oh okay

[stark crater]

it is from the video itself

I mean -1 x -1

the second -1 here

[rough furnace]

I'm not sure where that came from

[stark crater]

this seems to be the differentiation of whatever inside the parenthesis from numerator

[stark crater]

yeah.. it confused me too

[rough furnace]

Man this thing make no sense

where the x came from

[stark crater]

u mean dx ?

[rough furnace]

,w int (l+c-x)^(-2)

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yobi_fx.

[rough furnace]

No the x in the denom

[stark crater]

ommm..

[rough furnace]

btw you wrote the question wrong

[stark crater]

it is impossible to get x according to my problem

yeah I just noticed that, sorry

[rough furnace]

there was two -1 because this's -x

Not +x

[stark crater]

so this rule + the derivative of (x+c) in the denominator

[rough furnace]

and chain rule

Wel chain rule is quite painful to explain

You should watch youtube

[stark crater]

I know what is chain rule

u just do dertivative to what is inside the parenthesis

sorry for my english btw but I think u got what I said

[rough furnace]

okay okay

That's like everything you need I think

[stark crater]

yeah..

thanks alot

bye

.close

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[stark crater]

hello again

[stark crater]

using those integrations

I wanna find dE_y

that is my solution

however, I am not getting zero. my lecturer said that The result of E_y must be zero .. can anyone pls tell me where I have gone wrong ?

[neon iron]

U are missing a -ve sign

[neon iron]

It should be x-xp

Try to simplify maybe it will work

[stark crater]

om...

[neon iron]

U got it?

[stark crater]

x-xp ? why tho ?

[neon iron]

Look at your initial integra

The one u used as refrence

[stark crater]

oh u mean this ?

[neon iron]

Yes

Now in the ques (x-xp)^2= (xp-x)^2

And u have the formula for (x-xp)^2

I haven't simplified it but there's no mistake now so it should work

U got it?

[stark crater]

lemme try to solve it

but xp is a different thign

we are integerating in terms of x

and in that initial integral we should are integrating in terms of x too

so

we should have negative sign outside right?

because it is -x not +x

still.. I dont see any different that would lead to zero

[stark crater]

the first term is going to have L and last term cant have that L

how it will cancel each other

[neon iron]

No

Consider the situation of integral of (x-a)^2

[stark crater]

(x-a)^3 /3 ?

[neon iron]

According to you integral of( x-a)^2 and (a-x)^2 must be different but they are not
You can expand them and check

[neon iron]

That captures your point right?

[stark crater]

yeah..

I see..

[neon iron]

You got it?

Try to simplify

Maybe it will work

[stark crater]

yeah I will

thanks

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[shrewd birch]

also is there a better way to solve it?

seems like an easy soln but it took time fro me

bcoz what id do is make 4 pairs of x,y

<@&286206848099549185>

hi

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[shrewd birch]

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[silk carbon]

???

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[winter temple]

what is (f) about? Since we have the ordered basis $\beta={1,x,x^2}$, I see that $f(x):P_2(F)\to P_2(F)$, and $[f(x)]_{\beta}$ is the matrix representation $f(x)$ in the ordered bases $\beta$. But I couldn't figure out what $f(x)$ is.

[thorny flame]

afi

[hardy wing]

What is $\beta$?

[thorny flame]

Stitches

[winter temple]

$\beta={1,x,x^2}$ the standard ordered basis of $P_2(F)$

[thorny flame]

afi

[hardy wing]

so you want 3-6x+x^2 in terms of the elements of beta no?

[winter temple]

Umm, I thought $f(x)$ should be a linear transformation with a $3\times 3$ matrix representation?

[thorny flame]

afi

[hardy wing]

No that's T

You multiply f(x) (as a vector) by T (a transformation matrix), to get the output of the transformation (as a vector)

f(x) is a possible input to your transformation T

Distinct from T(f(x)), which would be the output of applying the transformation T on f(x)

[winter temple]

ohh

ig $[f(x)]_{\beta}=\begin{pmatrix}3\-6\1\end{pmatrix}$?

[thorny flame]

afi

[hardy wing]

yeah thats all

[winter temple]

alr thanks

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[arctic prairie]

proof for the image of an interval by a continuous function is an interval

[flat pike]

!status

[topaz sinew]

What step are you on?
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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

@arctic prairie Has your question been resolved?

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[calm charm]

But to find the $f (x)+g(x)$ graphic just add the ordinates for each x?

[thorny flame]

Reginald Puddingface

[void vapor]

I don't know why you said but at the start but yes

(this follows from how addition of functions is defined)

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[neat jasper]

what have you tried?

[undone walrus]

nothing. I don't understand

[neat jasper]

Okay

do you know what:
a) a topology is
b) an open set is
c) continuity is?

[undone walrus]

i do

[neat jasper]

all 3?

[undone walrus]

not continuity on topology tho

[neat jasper]

okay so

you want me to keep it abstract or show how it's based on continuity of real functions?

[undone walrus]

the simplest way pls

[neat jasper]

a function is continuous if the preimage of every open set is open

[undone walrus]

okay

[neat jasper]

so you need to check f^{-1}(S) for all open S, what are the open sets here?

[undone walrus]

{a}

?

[void vapor]

Those are not all the open sets

[undone walrus]

{b} and {a,b}

?

empty set and X ?

[void vapor]

{b} is not open

[void vapor]

And {a}, yes

[opal vault]

Here you need the definition of pointwise continuity

[undone walrus]

ok, so {a}, X and the empty set

[opal vault]

A function f:X-> Y is continuous at x if for any open set V that contains f(x), there is an open set U containing x such that f(U) subset V

I'll latex it one sec

A function $f:X\to Y$ is continuous at $x\in X$ if for any open set $V\subseteq Y$ that contains $f(x)$, there is an open set $U\subseteq X$ containing $x$ such that $f(U) \subseteq V$

[thorny flame]

Raphaelisius Maximus MMIII

[neat jasper]

no you don't need it

oooh

I take it back

[opal vault]

"Study the continuity at the points a and b of ..."

[neat jasper]

the question says "continuity AT"

yeah yeah youre 100% right

@undone walrus as you would expect, there's a theorem

a function is continuous iff it is continuous at every point

but it can be continuous at some points and not others

[undone walrus]

yes right

[neat jasper]

thank you for the correct Raphael

[undone walrus]

so we want {a} and {b} to be in V ?

[neat jasper]

V?

[undone walrus]

in this

this V

[neat jasper]

let's go through the 2 cases

[undone walrus]

okay

[neat jasper]

first you have f(a)=b

which means you want to consider every open set containing the element b

what are they?

[undone walrus]

X

[neat jasper]

right, that's the only one

is it open?

[undone walrus]

right

yes

[neat jasper]

so f(x) is continuous at x=a

where f(a)=b

[undone walrus]

ok perfect

[neat jasper]

now you do the other one

it's different of course

[undone walrus]

of so f(b)=a

there are two open sets containing a

{a} and X

[neat jasper]

right, so you need to look at

f^{-1}({a}) and f^{-1}(X)

the second one you already checked from the earlier part

wait actually i cheated

[undone walrus]

I did?

[neat jasper]

rather

I skipped a step you might not have gotten yet

so, pretend i didnt say it

😓

[undone walrus]

dont worry!

[neat jasper]

remember that inverse image distributes along union

so you can test f^{-1}({a,b}) = f^{-1}({a}) cup f^{-1}({b})

(regular image does not have this property)

(only inverse image)

let me write that in latex so there's fewer curly brackets

[undone walrus]

okay

[neat jasper]

$$f^{-1}({a,b}) = f^{-1}({a}) \cup f^{-1}({b})$$

[thorny flame]

gfauxpas

[undone walrus]

ok right

[opal vault]

Actually union is fine

[neat jasper]

did I mix it up with intersection?

[opal vault]

Yes

[neat jasper]

silly me

forward image distributes along union, but forward image does not distribute along intersection. inverse image distributes along both

[undone walrus]

of so f^-1 {X} = f^1 {a} union f^-1 {b} ?

[neat jasper]

good thing you're here!

[neat jasper]

yes because X = {a,b} here exactly

[undone walrus]

nice

[neat jasper]

is that open?

[undone walrus]

no?

[neat jasper]

what is it

what's the set

[undone walrus]

because it is {b} union {a}

so {a,b}

[neat jasper]

=X

[undone walrus]

ok so it is an open set

right?

[neat jasper]

indeed

now you have to check f^{-1}({b})

err

which one were we up to

b or a

f^{-1}({a})

[undone walrus]

f^{-1}({a})

exactly

[neat jasper]

=?

[undone walrus]

which is {b}

right?

[neat jasper]

open?

[undone walrus]

not open

[neat jasper]

so f(x) is not continuous at x=b

typo

by the way, if you want to think of why this is a definition of continuity

think of epsilons and deltas as open epsilon-balls and open delta-balls

there are open sets

[undone walrus]

for it to be continuos at x=b it would have to be open at both f^{-1}({a}) and f^{-1}(X)

[neat jasper]

correct.

[undone walrus]

right

[neat jasper]

these are*

they're not the ONLY open sets, but you'll learn why with real functions in the standard topology it's good enough to only check open balls

[undone walrus]

okay

thank you so much

[neat jasper]

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[merry helm]

how is it not

[torpid shard]

is it not what

[merry helm]

blue van

sorry had to think for a sec

it literally has the highest for all 3 months

[sharp pagoda]

its clearly violet

oh wait

[merry helm]

on what basis

correct answer is A (red)

idk why

[sharp pagoda]

actually not its green

oh wait no its red

i was trying to do this in my head lol

its not hard

[merry helm]

i dont get why you need to calculate anything, can't you just which one has the highest mileage in all 3 months

[torpid shard]

its the blue van

[sharp pagoda]

no

its the red

because its the end of month not start

[torpid shard]

this discussion is so stupid

Check on a month by month basis

[sharp pagoda]

why

its just november-august

[torpid shard]

Blue has always the highest amount of mileage no matter the month

[sharp pagoda]

huh

[torpid shard]

No combination of months could lead to other having more mileage

[sharp pagoda]

its asks for the greatest distance between september 1st and november 30th

[merry helm]

blue also has higher mileage in august anyways

[torpid shard]

If the blue van is not the answer in the problem, then the problem is wrong

[sharp pagoda]

no you're wrong

[torpid shard]

🥀

[sharp pagoda]

blue clearly has 83392-73959=9433

[torpid shard]

Oh, is this the general reading not the absolute amount

im so freaking dumb

lol

[sharp pagoda]

bruh

[torpid shard]

just search what month has the highest difference from august to november

[sharp pagoda]

red has 78853-68240=10613

[sharp pagoda]

thats what im trying to say bruhhhh

and you're telling me im wrong bruhhhh

and btw i already calculated that in head its red

[sharp pagoda]

here i said that before

[torpid shard]

when i looked that they were all increasing i figured its accumulating

[sharp pagoda]

i read the task normally.

[merry helm]

ahhhh

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[merry helm]

i don't understand how we can tell the opposites from the other die

just by looking at a different die

[ruby tree]

Well you can see the opposites on this die, can you list them?

(also why do they go back and forth between dice and die )

[sharp pagoda]

yea

[ruby tree]

@merry helm ?

[drifting swift]

i would try drawing nets of both dice and filling in based on the visible faces

oh wait hold on

this is ONE die you're looking at.

[ruby tree]

Yes it's completely determined

[drifting swift]

definitely draw a net for it and fill it in based on available info

take note of how the six-face is oriented

[ruby tree]

OP doesn't feel like answering though, so

[merry helm]

yeah

on the other die (which is shown)

3 and 1 are opposites, 4 and 2 are opposites, 6 and 5 are opposites

[merry helm]

sorry i was doing other maths while waiting

[merry helm]

but i dont really have any info on the other (not shown) die

i only have info on the die shown

[drifting swift]

so then you're saying the opposite-face pairs on the first die are 1-3, 2-4 and 5-6

those totals are 4, 6 and 11 so you cannot have any of the same totals on the second die

[sharp pagoda]

nor 7

[merry helm]

yeah this makes sense

would a fast way be to just check all of the available options

[sharp pagoda]

yea

[merry helm]

and see if those add up to the restricted totals

oke

[drifting swift]

forbidding total 6 gives you this right away

the third unmentioned pair also shouldnt add up to 4, 6 or 11

or 7

ok now actually im stuck too

D is out but i cant rule out any of the others

oh wait

[ruby tree]

You can't have the same sum on one die

[drifting swift]

C is out bc 2+6=3+5

A is out for the same reason

yeah

[merry helm]

but

why

8 is not a restricted total

right?

[drifting swift]

it's not restricted but in that option it ends up duplicated on the 2nd die

[sharp pagoda]

i think its easier to think about the sides of the dices

its ranging from 3 to 11, with 4, 6, 7, 11 out

and the sum must be 21

its not hard to see that its should be 3, 8, 10

[merry helm]

okay ty

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[opaque ridge]

How do i get the values of the roots?

I plugged it into desmos but idk how to set the intervals

[empty heron]

Can you show your desmos?

[opaque ridge]

yea

one sec

[empty heron]

You can click on the roots, no?

They’re called “point of interests”

[opaque ridge]

Yeah i put in those values but it said incorrect

[empty heron]

so it’s clickable

[opaque ridge]

because maybe its something about interval

(0,1)

[empty heron]

Not really uh maybe you need more sig figs

or like do you have any instruction

on how you should “approximate” your answer?

[opaque ridge]

this is the whole question

doesnt state about rounding or anything

wait

[gray mulch]

Can't you just zoom in to the required level, e.g. with the mousewheel?

[empty heron]

oh it says 4dp

[opaque ridge]

decimel rounded to 4

places

yeah i see that

[empty heron]

can u show what you wrote out?

[opaque ridge]

Wait lemme write it out again

[empty heron]

There’s no need to though

Those points are clickable

[opaque ridge]

Yeah this is what the roots are

after i clicked the points

But it says incorrect

[gray mulch]

0.14285 is not rounded to 0.1428

[opaque ridge]

oh yea

[opaque ridge]

there we go\

[empty heron]

Yeah your rounding is off

[opaque ridge]

got itright

thank you!

yeah that one number was off

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[crude quail]

This is how far I've gotten, and I wanna know whether there are any properties im missing? I can't see how else to solve this

[wintry meteor]

,rccw

[stiff knoll]

the abs function isn't linear: $\exists (a,b), |a+b| \neq |a| +|b|$

[crude quail]

Oh I see

But what about th triangle inequality?

It will be less than or equal to?

[stiff knoll]

less than or equal to yes

[wintry meteor]

Don’t think that’s useful here though

[thorny flame]

Médicis

[crude quail]

Oh

[stiff knoll]

you need to to analyse the function on $(-\infty , -1)$, $[-1, 1)$ and $[1, \infty)$

[thorny flame]

Médicis

[crude quail]

Why those values specifically?

[stiff knoll]

do you know the definition of the abs function?

[stiff knoll]

!occupied

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[crude quail]

The magnitude of difference between two points?

[stiff knoll]

no

[crude quail]

The magnitude of any point from 0?

[crude quail]

Ohhhh

Roght

[stiff knoll]

hence $|x+1| = x+1$ if $x+1 \geq 0$,

$=-x-1$ if $x+1<0$

[thorny flame]

Médicis

[stiff knoll]

therefore for |x+1| you've got to study the function on $[-1, \infty)$ and $(-\infty , -1)$

[thorny flame]

Médicis

[stiff knoll]

same thing for |x-1|

[crude quail]

How do those two relate?

[wintry meteor]

So when you add them together you have to treat where any term of the function can change separately

[crude quail]

I get up to this part

[wintry meteor]

$|x+1| = x+1$ if $x+1 \geq 0$,

$|x+1| =-x-1$ if $x+1<0\$

$|x+1| = x+1$ if $x\geq -1$,

$|x+1| =-x-1$ if $x<-1$

[thorny flame]

BBMaths

[crude quail]

Ohbhh

That makes sense

And then i would do the same thing for x-1?

How would one study these intervals though?

[stiff knoll]

well study them individually

they are distinct from one another anyway

[crude quail]

And then add them? And how does one study them?

[topaz sinew]

@crude quail Has your question been resolved?

[mint crescent]

Note that the definitions of the “parts” of $f$ change at $x=-1$ and $x=1$

[thorny flame]

Civil Service Pigeon

[mint crescent]

If you plot these on a number line, you should see how they split up the number line into three intervals: $(-\infty,-1)$, $(-1,1)$, $(1,\infty)$

[thorny flame]

Civil Service Pigeon

[mint crescent]

Obviously you need to include $x=-1,1$, but you can just throw those on wherever

[thorny flame]

Civil Service Pigeon

[mint crescent]

But my point here is that you can consider $f$ on each of these intervals

[thorny flame]

Civil Service Pigeon

[mint crescent]

Ex. For $(-\infty,-1)$, $|x-1|=-(x-1)$ and $|x+1|=-(x+1)$, so $f(x)=-(x-1)-(x+1)$ for that interval

[thorny flame]

Civil Service Pigeon

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[burnt swift]

Knowing that $a, b \in \mbb{Z}$ and $\gcd(a,b) = 5$, find the possible values of $$\gcd(a^4 + b^4, 2ab^2)$$ Show an example for each case

[thorny flame]

Renato

[topaz sinew]

@burnt swift Has your question been resolved?

[reef saddle]

This looks like a previous imo question am i wrong?

[burnt swift]

is a question from my intro to algebra class

[reef saddle]

Confused with this

[burnt swift]

alrighty

[quasi depot]

Hint, think what prime numbers can divide both 2ab^2 and a^4 + b^4

[burnt swift]

care to elaborate ?

[quasi depot]

Sorry I am busy

[burnt swift]

from the gcd(a,b) condition what I get is that

5 | a , 5 | b

from there we get that

5^4 | a^4

so

a = 0 (mod 5)
and so
a^4 = 0 (mod 5)

similarly with the 5 | b condition I get that 5^4 | b^4
so b = 0 (mod 5)
and we get b^4 = 0 (mod 5)

so b^4 = 0 (mod 5)

[burnt swift]

feel free to ping

because I got stuck with this one

[quasi depot]

Can 7 divide the gcd?

[burnt swift]

well

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@burnt swift Has your question been resolved?

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@burnt swift Has your question been resolved?

[mortal steeple]

5^4 | a^4
5^4 | b^4

so 5^4 | a^4 + b^4

and
5 | a
5^2 | b^2

so 5^3 | 2ab^2

so hcf is either 5^3 or greater for sure

@burnt swift

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[pure talon]

I need my work checked pretty pretty please

[ruby cove]

Looks correct

Good job

[pure talon]

YIPPEE

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[pure talon]

The textbook says it converges so I might’ve made an algebra mistake

[maiden plank]

you cannot cancel like this

[pure talon]

I FELL ASLEEP 😭

my bad

so what do i do?

do i split the fraction?

i mostly get an indeterminate form

so do i L'Hopital?

and then by L'Hopital, i get 1/0

which is 0

and since r is less than 1 bc r is 0 and 0<1, then that means it converges

?

[mortal steeple]

$\sum_{k=0}^\infty e^{-4} k = e^{-4} \sum_{k=0}^\infty k = -\frac1{12e^4}$

[thorny flame]

Wumpus Man

[mortal steeple]

[pure talon]

OHHH WAIT e TO THE -4 IS A NUMBER

i can move it out i forgot

howd u get -1/12

yk what i'll do it tmrw morning i have to wake up in 4 hours

[mortal steeple]

1 + 2 + 3 ... = -1/12

i dont remem tho

[maiden plank]

nononono

[mortal steeple]

im jk btw

[maiden plank]

oh ok

back to cave for me

[mortal steeple]

you should help, idk what convergence is

[maiden plank]

i mean

[pure talon]

https://tenor.com/view/lycoris-recoil-takina-takina-inoue-surprised-rocky-horror-picture-show-gif-26406896

im supposed to use ratio test

[outer laurel]

did you copy the problem correctly?

[pure talon]

lowk thats a good question i mightve been so sleepy i mightve copied wrong

[outer laurel]

that uh... says -k not -4

[pure talon]

WHAT

oh my god

[pure talon]

good catch 😭

im embarrassed

https://tenor.com/view/lycoris-recoil-takina-takina-inoue-surprised-rocky-horror-picture-show-gif-26406896

[maiden plank]

no wonder I couldn't figure out what happened

[pure talon]

my bad 😭

im actually going to sleep i cannot

.close

[topaz sinew]

Channel closed

Closed by @pure talon

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[spiral flint]

Any line passing through origin in R3 .. dimensions??

[spiral flint]

What is dimension?

[quasi depot]

Size of basis

Or equivalently the maximal size of a linearly independent subset

Or the minimal size of a spanning subset

[spiral flint]

So i can span with three basis?

(1,0,1)(0,1,0)(0,0,1)

[quasi depot]

!original

[topaz sinew]

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.

[quasi depot]

Span what?

[spiral flint]

Someone is asking me in dm

It's an old chapter for mr

I forget much of things

[quasi depot]

What is he asking you

[spiral flint]

@quasi depot

[quasi depot]

A line has dimension 1, a planr has dimension 2

[spiral flint]

And how?

Could you explain more please?

Ohh i have an idea

for equation in 3d

r=a+lambda b

So i can easily see that we need only one vector to represent a line in 3d

So dimension is 1

For a plane we need two vectors one is normal and one is situated over a plane

@quasi depot

n (r-r_0)=0

[quasi depot]

These are the dimensions of a line and a plane as subspaces of R^n

Need to go

[spiral flint]

Yes

[spiral flint]

Take care

.close

[topaz sinew]

Channel closed

Closed by @spiral flint

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[sick hawk]

hello everyone how are you all tonight

so for this question i tried this method

T X X C X X X X X

X = Letters that are not T or C

[quartz atlas]

Did you get the answer

[sick hawk]

nah im asking why my methode is incorrect

method

[fallow igloo]

[quartz atlas]

T X X C X X X X X

Here X can be anything like you said , but we can move TXXC as whole throughout the word

[sick hawk]

ok so C T can be swapped around
i choose 9 - 3(e) - 1(t) - 1(c)
so 2! * 4P2 * 6!/3!(three e)

[fallow igloo]

no one says the other letters cannot be on the left side

[sick hawk]

yeah i considered this

[sick hawk]

and i considered this

[sick hawk]

this is what i did

6! for the TXXC

and the other 5 letters

5P2 for the 2 letters between the T and C

and hello sky and night

[quartz atlas]

Ok so did you get the answer by the method you said