help-17

[spiral turtle]

If you found a pair (a, b) wouldn't (b, a) also work?

Unless a = b

[dark kiln]

it's way more

[surreal marten]

2009 nd 1

[dark kiln]

x from 1 to 2008, means 2008 pairs

[surreal marten]

thq question is x^2

and y^2

[spiral turtle]

Yes

[surreal marten]

i forgot to add sign

[spiral turtle]

It was understood

[surreal marten]

how did u do this could u give me steps

[spiral turtle]

Factor the original expression

[dark kiln]

i don't know how to factor

[surreal marten]

its my doubt dude

[surreal marten]

could u send me steps if possible

[spiral turtle]

We see that x^2 + y^2 + 2xy appears in the expression, so we can reasonably conclude that the form is probably (x + y + a)(x + y + b)

Then we multiply out this and figure out what a and b will work

[kind light]

why did i randomly think this didnt work :p

[surreal marten]

k .close

.close

[vocal sleet]

Channel closed

Closed by @surreal marten

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[knotty stream]

i need help in funcations grade 11

[knotty stream]

anyone can help

[odd helm]

Okay politoad guy send the question then

[knotty stream]

ok

[odd helm]

Complete the square/ write it in vertex form depending on where you are from

[knotty stream]

ok

[vocal sleet]

@knotty stream Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[wide marsh]

can someone help me how can i find the expression of the graphique one ?

[wide marsh]

i mean the first graph

[vocal sleet]

@wide marsh Has your question been resolved?

[wide marsh]

@graceful ibex

<@&286206848099549185>

[gaunt sparrow]

Well the main thing is that we don't know what the base is.

So if the function has the form $a^x$, can you use the point you have, $(-1,6)$, to solve for the base $a$?

[twin meteor]

Azyrashacorki

[wide marsh]

idk

all i know is that need to be smth like (1/3)^x

since its an - exponentielle

[gaunt sparrow]

With the point given, we know it should be $f(-1) = a^{-1} =6$. Can you solve for $a$ here?

[twin meteor]

Azyrashacorki

[wide marsh]

idk how to solve it

[gaunt sparrow]

$a^{-1} = \frac{1}{a} = 6$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

What about now?

[wide marsh]

is that going to give me 1/6?

[gaunt sparrow]

Yes.

So $a=\frac{1}{6}$

[twin meteor]

Azyrashacorki

[wide marsh]

but it aint touching the y in 3 with that

[gaunt sparrow]

Oh yes mb we should've done that first.

Tu as raison qu'on devrait ajouter un + 2 après, donc la forme générale devrait être $f(x) = a^x +2$.

Maintenant on peut refaire le tout en considérant le point (-1,6) à nouveau.

Ça nous donne $\frac{1}{a} + 2 = 6$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

De là tu peux trouver la valeur de a et substituer dans la forme avec laquelle on a débuté

[wide marsh]

et pk 2 et pas un autre chiffre?

ca me donne 1/4

mais c'est toujours pas bon

[gaunt sparrow]

Merde je fais vraiment du n'importe quoi.

Puisque l'asymptote à y=0 est toujours là c'est plutôt un coefficient qu'on doit ajouter au devant.

Du coup, on a $f(x) = b a^x$.

En substituant x=0, on obtient $f0) = b = 3$.

Ensuite on peut réécrire sous la forme $f(x) = 3a^x$ et utiliser ça avec le point (-1,6) pour trouver la base.

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Ça devrait marcher cette fois 😅

[wide marsh]

je t'avoue que je suis perdue je ne sais pas comment le resoudre la

[gaunt sparrow]

Avec le point on retrouve l'équation $6 = 3\frac{1}{a}$.

[twin meteor]

Azyrashacorki

[wide marsh]

je comprends tjrs pas 😅

[vocal sleet]

@wide marsh Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[vital latch]

How do I calculate the limit as x goes to infinity for sqrt(x^2+x) - sqrt(x^2-x)?

[vital latch]

My idea was to factor out x in both and get

Sqrt(x(x+1)) - sqrt(x(x-1))

Then say that we have (x^1/2 * (x+1)^1/2) - x^1/2 * (x-1)^1/2

Then factor out x^1/2 from that and get

X^1/2((x+1)^1/2 - (x-1)^1/2)

But then i kinda get stuck

[dense cedar]

Multiply with conjugate

[vital latch]

ah

Yeah will try that - cheers

[dense cedar]

Then divide numerator and deno with 1/x

And put 1/x = 0

[vital latch]

Tyty

.close

[vocal sleet]

Channel closed

Closed by @vital latch

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[weak meteor]

ok so basically, I have a vector whose components are all random picked from a gaussian distribution, (making a sphere of possible vectors i guess), and I am trying to figure out how to stretch this "sphere" along a (normalised) vector. I have tried just multiplying the elements of the random vector by the elements of the stretch direction vector, but that doesn't seem to work, any ideas?

[weak meteor]

I would guess there is funny linear algebra stuff needed here, but I have way too little experiance to work this out myself

[peak matrix]

If i'm not mistaken, all vectors would be possible

also, vectors in R^3?

[weak meteor]

yep

[peak matrix]

.

every value is possible in gaussian distribution, those that are far from the center are just extremely unlikely

[weak meteor]

you have a point, but from what I've seen of the code I'm using to generate the guassian, there's definitely limits on it.

[peak matrix]

there aren't limits

the limits are imposed by computer and time

if you kept it running for a long enough time on a computer that can deal with extremely small probabilities, you would most likely generate any vector

[weak meteor]

so is this like, a reason I can't change the shape of it? or what?

[peak matrix]

the shape wouldnt be a sphere

it would be a whole space

[weak meteor]

well yeah, Im not wanting a sphere

i want it to be like a gaussian, but stretched along a vector

wait

when I said "sphere" i was more meaning like how a gaussian is symmetrical, not that it actally was a sphere

so thats just me being a dingus, sorry about that

[peak matrix]

you want even get anything like gaussian

all vectors are possible

hence the "shape" of all possible vectors is just the space itself, R^3

[weak meteor]

ah

[peak matrix]

on a computer, you cant generate all of them though

you can generate at most finitely many of them

and even if you kept the computer running for eternity, you would only generate countably many of them

while there are uncountably many

[weak meteor]

i see

i think what Im trying to say is that instead of a gaussian, which is symmetrical or whatever, I'm needing a way to scale that somehow, so that points are more likely along the stretch vector

[peak matrix]

so you got a normalized vector along which you'll be stretching

[weak meteor]

yep

[peak matrix]

say we wanna stretch red along black

firstly we project red into red

so we find the component of red that's parallel to black

then we find the perpendicular component

and decompose red to those 2 components

perpendicular component will be unaffected

parallel component will get multiplied

then add them back

and you get your stretched vector

[weak meteor]

so how would I do that in 3d? since the parallel's easy enough to find, but how would I get the two perpendicular components?

[peak matrix]

it's sufficient to find 1 perpendicular

v - p, where p is the projection of v

then p + (v - p) = v

so it will be a correct decomposition

[weak meteor]

i see

[peak matrix]

do you know how to do the projection part?

[weak meteor]

something about dividing dot products by lengths, but it would probably be best if you could tell me

[peak matrix]

this one is better

projection of a onto v = that

we actually know that |v| = 1

so you can leave that part out

since v is normalized

so it will be (a * v) v in your case

[weak meteor]

nice, thanks!

[peak matrix]

np

[vocal sleet]

@weak meteor Has your question been resolved?

Channel closed

Closed by @weak meteor

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[jaunty gorge]

I am completely lost for this

[hybrid flicker]

which part troubles you?

[jaunty gorge]

in the second "lim"

because I don't really see where the sqrt come from in the denom

[hybrid flicker]

square root and square cancel each other

$x = (\sqrt{x})^2$

[twin meteor]

rafilou2003

[jaunty gorge]

Ohhhh true I keep forgetting there's multiple work arounds for these stuff

gotta think outside the box

thanks, I appreciate it

.close

[vocal sleet]

Channel closed

Closed by @jaunty gorge

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>.
• 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.

[real furnace]

I get everything until the part where it says "a>2" on the second to last line

Why does it say a>2?

Oh wait nvm

Why is -3a<-3(2)

[vocal sleet]

@real furnace Has your question been resolved?

[vocal sleet]

@real furnace Has your question been resolved?

[lone linden]

Multiply both sides by -3 and recall that when you multiply both sides of an inequality by a negative number, you reverse the direction of the inequality sign

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[fickle oasis]

help

[vocal sleet]

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

[fickle oasis]

4

imma get a pic of my work brb

is this right?

I think I might have got it wrong earlier because I divided the whole thing by 15 instead of 16

stupid me

[sly sierra]

i get the same answer as you

[fickle oasis]

it should be right then I got an exact answer this time at least xD

.close

[vocal sleet]

Channel closed

Closed by @fickle oasis

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[untold arrow]

Given the differential equation ( y'' + p(x)y' + q(x)y = f(x) ), which has three solutions ( y_1 = x ), ( y_2 = e^x ), and ( y_3 = e^{-x} ), find the particular solution ( y ) that satisfies the initial conditions ( y(0) = y'(0) = 3 )

[twin meteor]

riyobi

[hybrid flicker]

!status

[vocal sleet]

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

[untold arrow]

1

[hybrid flicker]

ok so do you know how to find all the solutions of a linear ODE?

in general

[untold arrow]

1. Find the integrating factor, μ(x), which is given by μ(x) = e^∫P(x)dx.
2. Multiply both sides of the ODE by the integrating factor, μ(x), to get μ(x)dy/dx + μ(x)P(x)y = μ(x)Q(x).
3. Recognize that the left-hand side is now the derivative of μ(x)y, so we can rewrite the equation as d(μ(x)y)/dx = μ(x)Q(x).
4. Integrate both sides with respect to x to get μ(x)y = ∫μ(x)Q(x)dx + C, where C is the constant of integration.
5. Finally, divide both sides by μ(x) to get the general solution, y = (1/μ(x))(∫μ(x)Q(x)dx + C).

[hybrid flicker]

did you use an AI for that?

[untold arrow]

Yes, since the steps are a lot so I didn't type myself

[hybrid flicker]

ok but

for example

what do you know about the set of solutions of an homogeneous ODE?

[hybrid flicker]

for example if f(x) = 0 here

[untold arrow]

Hmm

Idk because they didn't give the general solutions of the equation

[hybrid flicker]

they gave 3 solutions to the equation

but what can you say about the set of solutions of a homogeneous linear ODE?

(part of the answer is in the question)

[untold arrow]

y1 and y2 are the general solutions of the original equation?

[hybrid flicker]

the general solution encompasses ALL possible solutions

I bet that I can create another solution that differs from the 3 we've been given

such as x - e^x + e^(-x)

not really no

c1 and c2 can't be any real number

and that doesn't yield all solutions

let's just forget about y1,y2 and y3

[hybrid flicker]

and focus on this for now

[untold arrow]

the solutions to the homogeneous equation are y1 y2 y3

[hybrid flicker]

no

homogeneous means "f(x) = 0"

[untold arrow]

I'm confused

y'' + p(x)y' + q(x)y = 0

[hybrid flicker]

ok so

solutions of y'' + p(x)y' + q(x)y = f(x) are just solutions of the original ODE

solutions of y'' + p(x)y' + q(x)y = 0 are solutions of the homogeneous ODE

[untold arrow]

Ooh

[hybrid flicker]

what do you know about the **space ** of solutions to a linear homogeneous ODE?

[untold arrow]

Iirc : [ y(x) = c_1y_1(x) + c_2y_2(x) + \ldots + c_ny_n(x) ]

[twin meteor]

riyobi

[hybrid flicker]

so for a homogeneous ODE

you can just add linear combinations of the solutions you find

and get another solution

sounds like a vector space

[untold arrow]

Yes

[hybrid flicker]

alright

and one more thing

once we have all the homogeneous solutions

[hybrid flicker]

to find the general solutions

all we have to do is pick 1 solution

we call it y_p

(p for particular)

and then

all the other solutions are given by

$y_c = y_p + y_h$

[twin meteor]

rafilou2003

[hybrid flicker]

(h stands for homogeneous)

[untold arrow]

Yes

[hybrid flicker]

welp

we already have one particular solution

(x for example)

we just need the homogeneous solutions

to do that... what happens when we subtract y_1 to y_2 for example

well both of them verify y'' + p(x)y' + q(x)y = f(x)

so y_1 - y_2 is gonna verify

y'' + p(x)y' + q(x)y = ... 0

we have a first homogeneous solution

since this is a 2nd order ODE

we only need 2 independent ones

and then we have our homogeneous solution space

to get a second one

well y_1 - y_3 for example

same thing

so $y_h = c_1(x-e^x) + c_2(x-e^{-x})$

[twin meteor]

rafilou2003

[hybrid flicker]

we add our particular solution in the bunch and...

$y_c = x + c_1(x-e^x) + c_2(x-e^{-x})$

[twin meteor]

rafilou2003

[hybrid flicker]

got it?

[untold arrow]

yes

[untold arrow]

Then we need to write down other cases when the particular solution is the other two?(e^x or e^-x i mean

[hybrid flicker]

no it results to the same thing

whatever particular solution you pick

[untold arrow]

Ohh ok

Darn my internet delayed so bad

But then what to do the next ?

Plug it in?

( y'' + p(x)y' + q(x)x + q(x)(c_1(x-e^x) + c_2(x-e^{-x})) = f(x) )

[vocal sleet]

@untold arrow Has your question been resolved?

Channel closed

Closed by @untold arrow

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[inner osprey]

i assume this should say g'(x) never equals 0 rather than g'(x) ≠ 0

[untold arrow]

Aren't both expressions basically the same?

[vocal sleet]

@untold arrow Has your question been resolved?

[vocal sleet]

@untold arrow Has your question been resolved?

[vocal sleet]

@untold arrow Has your question been resolved?

[vocal sleet]

@untold arrow Has your question been resolved?

[granite flicker]

@untold arrow this is a wild one and took me a while

but check this out

Notice that the statement $g'(x) \neq 0$ means that g must be strictly monotonic (Darboux's theorem). I will attempt to prove this by contradiction, first suppose that such $\xi$ doesn't exist.
then $f'(x) \neq 0$ as well making f(x) monotonic. then for $\int_a^b f(x)dx = 0$ to be true there must exist a unique $c \in (a,b)$ such that $f(c) = 0$. Now observe this equivalence:
[
\int_a^b f(x)dx = 0 \leftrightarrow \int_a^c |f(x)| dx = \int_c^b |f(x)|dx
]
which is true because f(x) has the same polarity before c and opposite polarity after c.
Now define $u(x) = \alpha + \beta g(x)$, such that $\beta >0$, and $\alpha,; \beta$ are chosen to make $u(x) > 0, ; u(c) = 1$ then depending on the type of monotonicity you would have either
[
\int_a^c |f(x)g(x)| dx < \int_a^c |f(x)| dx = \int_c^b |f(x)|dx < \int_c^b |f(x)g(x)|dx
]
If $g(x)$ was monotonically increasing, and
[
\int_a^c |f(x)g(x)| dx > \int_a^c |f(x)| dx = \int_c^b |f(x)|dx > \int_c^b |f(x)g(x)|dx
]
If it was monotonically decreasing, and since $u(x)$ is positive then $f(x)g(x)$ has the same polarity before c and opposite polarity after c. hence
[
\int_a^b f(x)u(x) dx \neq 0
]
but if that's the case then
[
\int_a^b f(x)g(x) dx = -\frac{\alpha}{\beta}\int_a^b f(x) dx +\frac{1}{\beta}\int_a^b f(x)u(x) dx = \frac{1}{\beta}\int_a^b f(x)u(x) dx \neq 0
]
thus finishing the proof by contradiction.

[twin meteor]

Mohamed Mohsen

[untold arrow]

you just gave the full solution?

[granite flicker]

honestly, with these type of questions, I don't even know what hint to give.

[vocal sleet]

Channel closed

Closed by @untold arrow

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[smoky pebble]

i solved it already but it says that i cant directly use manipulation to solve here

[vocal sleet]

Please don't occupy multiple help channels.

[smoky pebble]

using x = 1/t

[rare swift]

Wdym by direct manipulation?

[smoky pebble]

write it as 1/y-ln(1+y)/y.1/y

comes out to be 0

[rare swift]

Did u try standard limits

Divide and multiply the expression by 1/x

U will get x - x(log(1+1/x) /1/x
So you will get x - x = 0

Since log(1+x) / x where x tends to zero = 1

[smoky pebble]

no i did it using substitution

i got right answer

but it was written in remarks that that method will give wrong answer

so i wanna know why so

[rare swift]

it won't

Rather it shouldn't
What the actual answer

[smoky pebble]

it is written clearly

in my book

that it will give wrong answer and it illustrates it too

[smoky pebble]

1/2

[smoky pebble]

clearly gives 0

[rare swift]

But we can't think of a flaw in their argument too

[smoky pebble]

must be something complicated

[rare swift]

[smoky pebble]

it didnt explain just said

it gives wrong answer

[rare swift]

Try taking the remainder of Taylor expansion

[smoky pebble]

?

[rare swift]

The n+1 fold integral try taking it for 3rd term

[smoky pebble]

i dont know what it is tbh

not that good at math 😭

i mean i know some integration but i dont know what fold is

[rare swift]

Have u studied taylors theorem?

[smoky pebble]

nope just know taylor series expansions

[rare swift]

Ahh k then leave it

It's complicated tbh

Still don't know where that 1/2 came from

[rare swift]

Plot it on desmos
Limit does not exist

[smoky pebble]

expand the series

and cancel out after taking common

[rare swift]

Yea i can see how they got 1/2 I'm saying I don't understand how 1/2 has any significance on the graph

[surreal basin]

red is y = x - x^2 ln(1 + x)
blue is y = 1/2

[rare swift]

Can't send a pic cause it isn't loading

[rare swift]

yea so limit dosent exist

[surreal basin]

how

[rare swift]

Why does 1/2 have any significance

[surreal basin]

thats the horizontal asymptote of x - x^2 ln(1 + x)

[rare swift]

It approaches infinity from one side and zero from the other

[surreal basin]

show me those sides then

remember this is a horizontal asymptote

[rare swift]

Ohh wait fck

[surreal basin]

jfc

[rare swift]

Nvm sry sry I'm high

[rare swift]

Then why do we get zero by standard limits

[surreal basin]

you messed up

thats the only reason

[rare swift]

How
Possible taking into account I'm high

[rare swift]

U do it

[surreal basin]

can you walk through your thinking in saying that this limit is 0 instead of indeterminate

[smoky pebble]

oh nah

i get it now

it meant to remark that

u cant partially apply

limits

[rare swift]

Take the standard limit ln(1+x) /x is 1
We get lim(1/y) - lim(standard limit) ×lim(1/y)
Which both cancel out to get 0

[smoky pebble]

i got it

nah u cant partially aplly limits

[surreal basin]

yea AI's correct

[smoky pebble]

yea i get it

[surreal basin]

you cant just apply one limit and then not the rest

[smoky pebble]

lol

[surreal basin]

once you apply that limit in the middle, you get (undefined) - 1 * (undefined)

limits are applied from the outside in, not the inside out

[rare swift]

Ohh wait
Fck

[surreal basin]

after I reremember ln(1+x), I do have the way to solve this problem using taylor series

[rare swift]

huh?

[surreal basin]

its this, right?

[smoky pebble]

yea

[surreal basin]

nice

[smoky pebble]

i did that

only

[rare swift]

Yea if u take till n dont forget to add the remainder

[rare swift]

Yea that's what the book has done

[vocal sleet]

@smoky pebble Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[tawny nacelle]

hmm, if I have a Hausdorff space X, can removing a point from it remove its Hausdorff property?

[tawny nacelle]

say, if X = S^1, and I delete one point... does the modified circle remain Hausdorff?

[atomic forum]

yes

[tawny nacelle]

hm... so it's true in general?

if X is Hausdorff, then X \ {p} is Hausdorff, where p \in X?

[atomic forum]

yes, i think so

[tawny nacelle]

[atomic forum]

(i forget if the empty space is Hausdorff)

[dusty ice]

it is a fact that subspaces of a hausdorff space are themselves hausdorff

[atomic forum]

mhm

[tawny nacelle]

ah, right

hmm

[tawny nacelle]

it is

[atomic forum]

then yeah

[tawny nacelle]

you remember how I wanted to show that the wedge sum of Hausdorff spaces was Hausdorff?

I think I can break it into a few cases

if if x and y are in the same Hausdorff space, then we're done by assumption

if x is in X_a but y is in X_b, then we can take the open subsets to be X_a and X_b \ {p_b}, where {p_b} is the base point for X_b...

I think?

[dusty ice]

X_a isn't open in the wedge sum though

[tawny nacelle]

ah, it's not?

[dusty ice]

no, because its preimage in the disjoint union is X_a and then a single point in each of the other spaces

this is not in general open

[tawny nacelle]

okay then let's just take X_a \ {p_a} then

now...

one more case

if either x or y is a base point...

WLOG, assume it's x

in this case... what should we do?

the open nbhd of y can stay as X_b \ {p_b}

but what about the open nbhd of x?

hmm

[atomic forum]

ok so it’s like

if you use the universal property of the wedge sum

then open subsets correspond to picking an open subset in each space, such that either all of them contain the basepoint, or none of them do

maybe this will help

[tawny nacelle]

hmm

[dusty ice]

in other words the open sets containing the base point are precisely those that are open in each space you've wedged, separately

[atomic forum]

you can figure this out by considering the continuous maps from the wedge sum to the sierspinski space

[tawny nacelle]

hold on

[dusty ice]

sierpinski is not a word I was expecting to come up in this discussion

[atomic forum]

the sierpsinski space is the classifying space for open subsets

[dusty ice]

yes but still

[atomic forum]

and then the universal property of the wedge sum lets you figure out what continuous maps out of the wedge sum are

[tawny nacelle]

okay, I'm back

so I need to pick an open nbhd of x that's disjoint with X_b \ {p_b}

[atomic forum]

that’s gonna be hard if x is in X_b lol

[dusty ice]

(x is in X_b in this case)

[tawny nacelle]

wait, what?

didn't I resolve that case already?

[tawny nacelle]

.

[atomic forum]

ok how exactly are you done

[tawny nacelle]

well, if x and y are both in X_a, then we can find open nbhds U and V around x and y st U \cap V = \emptyset, since X_a is Hausdorff, no?

this doesn't work?

[dusty ice]

U, V open in X_a yes,

[tawny nacelle]

oh.

[atomic forum]

but you need it to be open in the wedge sum

[atomic forum]

.

[tawny nacelle]

they're not necessarily open in the wedge yeah

I see

[atomic forum]

that’s what open subsets in the wedge sum look like

so you’ll be in trouble if U or V contain the basepoint

[tawny nacelle]

there must be some way to remove the base point then

[dusty ice]

(unless x or y are the basepoint)

[atomic forum]

mhm, but then you have to modify your open subset to make sure it’s actually open in the wedge sum

[tawny nacelle]

well yes, but I am dealing with that case separately

or does it make more sense to do it together?

instead of breaking this into so many cases?

[atomic forum]

maybe try understanding this first

it’s gonna be hard to prove the wedge sum is Hausdorff if you don’t know what open subsets in it look like

[tawny nacelle]

the open subsets of the wedge?

[atomic forum]

yes

[tawny nacelle]

hm

[atomic forum]

it makes some sense with the geometric picture

and/or you can use the universal property

[tawny nacelle]

okay let's wedge 2 circles

this is the case we showed earlier, when x and y belong to different spaces and neither is the basepoint

[atomic forum]

mhm

[tawny nacelle]

now we have this case

what is an open subset of the wedge going to look like?

[atomic forum]

.

[tawny nacelle]

yes, but visually

[atomic forum]

i mean, if you know how to pick open subsets in the circle visually…

they’re unions of open balls

[tawny nacelle]

is this an open subset of the wedge?

[atomic forum]

yes

you’ve picked an open subset from both circles

one is nonempty, and one is empty

and neither contains the basepoint

[tawny nacelle]

why does an open subset in the wedge need to contain stuff from both spaces?

I don't think I understand why

[atomic forum]

.

[tawny nacelle]

that's not what I meant

[atomic forum]

then i don’t understand what you mean

the subset you drew is open

[tawny nacelle]

hmm, hold on

[atomic forum]

you’ve picked a nonempty subset of the top circle

and an empty subset of the bottom circle

both are open

and both do not contain the basepoint

[tawny nacelle]

nvm I answered my own question lol

I was thinking of it the wrong way

okay wait, if it does contain the base point, then what happens?

[atomic forum]

if any of the open subsets you pick contain the basepoint, then all of them have to

[tawny nacelle]

what if some of them don't?

[atomic forum]

then it won’t be open in the wedge

[atomic forum]

yes that’s open

you’ve picked nonempty open subsets of both the bottom and top circles

and both contain the basepoint

[tawny nacelle]

wait, what?

how?

[atomic forum]

otherwise I’m confused at your drawing

[dusty ice]

you've drawn the lines too big and are hiding what I think you're trying to show

[atomic forum]

what subset of the top circle have you picked?

[tawny nacelle]

the top circle contains the base point

[dusty ice]

you meant this? (with the top bit there as well)

[tawny nacelle]

the bottom circle is this one

mhm

[atomic forum]

so the subset of the top circle you’ve picked is only the base point?

[dusty ice]

is this open though? what ball around the wedge point is entirely in the open set?

[tawny nacelle]

wdym only the base point?

[tawny nacelle]

none

[atomic forum]

i am asking you what subset of the top circle you are picking

[tawny nacelle]

let me colour it

[atomic forum]

ok then this fails to be open in the wedge

[tawny nacelle]

yes, but I wanted to ask why

[atomic forum]

the red set seems to be open in the top circle, and contains the basepoint

the blue set is open in the bottom circle, and does not contain the basepoint

[tawny nacelle]

I get that

[atomic forum]

so it fails the condition i laid out

[tawny nacelle]

but why does this mean it's not open?

[atomic forum]

in this case the wedge sum is metrizable

[dusty ice]

and no open ball around the wedge point is entirely in the open set you've chosen

[tawny nacelle]

but I don't see why the condition is tied to the openness of the set

[atomic forum]

so you can also think in terms of open balls

[atomic forum]

so you can use the definition of the quotient topology

[tawny nacelle]

that makes sense

[atomic forum]

and/or the universal property

[dusty ice]

so it's not open by definition, because in this case the drawing with metric is the wedge sum (up to homeomorphism)

[atomic forum]

yes, though i guess you’d have to prove that

the topology on the wedge sum is a quotient topology

[tawny nacelle]

so for U in the wedge to be open, we need q^-1(U) to be open?

[atomic forum]

the topology on this drawing comes from it being a subspace of R^2

[atomic forum]

yes, that’s the definition

[atomic forum]

though I slightly prefer this

[tawny nacelle]

if U is the union between the red and blue parts I drew earlier, which part becomes not open under the preimage

[atomic forum]

the preimage is not open

in the disjoint union topology

[dusty ice]

the red includes the wedge point, so in the preimage you have a single point in the bottom circle

[tawny nacelle]

right...

[atomic forum]

yes, and that together with the other set in the bottom circle is not an open subset of the bottom circle

[tawny nacelle]

wha

[atomic forum]

so it’s not open in the disjoint union topology

[tawny nacelle]

why's the blue part not open in the bottom circle?

[atomic forum]

that’s not what I said, higher

[tawny nacelle]

oh huh

[atomic forum]

i said that the blue part plus the basepoint is not open in the bottom circle

[tawny nacelle]

I see

that makes sense

I agree

[atomic forum]

so the preimage fails to be open in the disjoint union topology

[tawny nacelle]

right, okay

[atomic forum]

quotient topologies can be hard to grasp

[tawny nacelle]

I can tell

so I need to pick nbhds of x and y that do not contain the base point?

[atomic forum]

yeah

this is why I really think you should try to understand the universal property

[tawny nacelle]

I don't yet know the universal property for quotients (though I know you wrote stuff about it in my thread)

[atomic forum]

mhm, i think it makes working with them a lot easier

it definitely did for me

[tawny nacelle]

I guess

but I do want to do this exercise without that property too

I'm pretty sure Lee "intended" me to do it w/o universal properties

[dusty ice]

an alternative to the universal property is through saturated sets

[atomic forum]

sure

i don’t necessarily agree with that but that’s ok

[tawny nacelle]

I should probably mention that I don't yet know what those are either

I know they're coming up though

[atomic forum]

the wedge sum of two circles is metrizable

so you can use metric space intuition to help

[tawny nacelle]

well, I'm supposed to prove this for general topological spaces...

[dusty ice]

"intuition to help"

[atomic forum]

that doesn’t mean you’re not allowed to use metric space intuition

[tawny nacelle]

ah, true

[atomic forum]

it is very useful for lots of topology problems

[dusty ice]

most topologists only think about R^2

[atomic forum]

like your first instinct should really be to draw a picture

not work with the symbols

if you try to approach topology the same way you’d approach, like, algebra

you won’t have a fun time

[tawny nacelle]

I have a few pictures I suppose

I'm trying to figure out what open nbhds to take though

oh hold on

I can choose X_a \ {p_a, y} to be the open nbhd of x, and X_a \ {p_a, x} to be the open nhbd of y, yeah?

@atomic forum @dusty ice

[atomic forum]

yep

[tawny nacelle]

okay, nice

now for the case where we have x as a basepoint, and y is whatever

[atomic forum]

mhm

[dusty ice]

these intersect in general?

[tawny nacelle]

oh uhh

[atomic forum]

oh yeah true, so you’d have to worry about that

[tawny nacelle]

I can probably tweak this

lemme see

[atomic forum]

you should know that compact subsets of Hausdorff spaces are closed

in particular finite subsets are always closed

[tawny nacelle]

okay hold on

we know that X_a is Hausdorff, so we can choose open nbhds U and V that are disjoint in X_a

[atomic forum]

mhm

[tawny nacelle]

then... can I remove p_a from each of these sets and call it a day, or is it possible I lose the openness of U and V when I do so?

[atomic forum]

well, how would you check whether U \ {p_a} is open?

[tawny nacelle]

well, its preimage is going to be entirely contained in X_a, since it doesn't have the base point

[atomic forum]

i mean open in X_a

[tawny nacelle]

oh

well, we need to find an open nbhd around each point in U \ {p_a}

that's contained in U \ {p_a}, that is

okay, I'll come back to this case later...

in the case where x is a base point and y is anything else, we have smth like this

can I just take X_a (the top circle) to be the open nbhd of x, and X_b \ {p_b} (bottom circle minus base point) to be the open nbhd of y?

they're disjoint, and if I haven't made a mistake, they should be open in the wedge also?

I'll be back in 15 min

[atomic forum]

X_a won’t be open in the wedge sum

because what you’re doing is choosing X_a from the top circle and the empty set from the bottom circle

these are both open in their respective spaces, but one contains the basepoint and the other does not, so it won’t be open in the wedge sum

[vocal sleet]

@tawny nacelle Has your question been resolved?

[tawny nacelle]

I see

then could I pick X_a \cup {p_b} to be the open nbhd of x?

oh, but {p_b} isn't open in X_b, is it?

hm

okay wait, since X_b is Hausdorff, we can separate p_b and y with open sets U and V

then take X_a \cup U to be the open nbhd of x

and V to be the open nbhd of y

these are disjoint, X_a \cup U is open in the wedge since they contain the base points, and V is open in the wedge cause neither it nor the empty set contain the base points

[atomic forum]

Yep!

You’re getting the hang of it

[tawny nacelle]

but hm..

do I need to deal with a new case now: x = p_a, and y = p_b?

[atomic forum]

Those are the same point in the wedge

[tawny nacelle]

oh no

yes, you are right

I agree

so now I only have to deal with x,y in X_a, and neither is a base point

[atomic forum]

The only case you’ve left out is when x and y are in the same space, but neither is the basepoint

Yeah

[tawny nacelle]

I feel like U \ {p_a} and V \ {p_a} should work, but I am not 100% confident

well, let's see

these are obviously disjoint cause U and V are disjoint

U \ {p_a} should be open in the wedge cause neither it nor the empty set contain their base points

and likewise with V \ {p_a}

this surely works... yeah?

[atomic forum]

Why is U\ {p_a} open in X_a?

[tawny nacelle]

hm, good point

[atomic forum]

Once you’ve got that i think you’re done

[tawny nacelle]

well, not exactly, right?

[atomic forum]

Isn’t this the last case

[tawny nacelle]

yes, but shouldn't I prove that a subset of the wedge is open iff either its intersection with each X_a contains X_a's basepoint, or it doesn't?

[atomic forum]

Oh you mean - prove the condition i stated

[tawny nacelle]

yes

[atomic forum]

Sure yeah

Ig i used the universal property to deduce that condition, but you’d have to prove it

[tawny nacelle]

okay, I've got some work to do then

[atomic forum]

And/or you can just check that all the subsets you use are open by directly using the def of quotient topology

[tawny nacelle]

ah, true

[tawny nacelle]

I've got to go have dinner soon, but I'll try to first prove this when I come back

[atomic forum]

There’s a nice way to do this for arbitrary quotient topologies btw

[tawny nacelle]

before the condition thing

[atomic forum]

Open subsets of X/ R correspond to open subsets of X which are closed under R

So you require that if x in U and x R y, then y in U

[tawny nacelle]

ic

okay, I'll be back in a bit

thank you for all the help Pseudo (and Edward, if you're still here)!

[atomic forum]

im here though im gonna head to sleep soon

[vocal sleet]

@tawny nacelle Has your question been resolved?

[tawny nacelle]

ah, no worries then

I can write up and post in my thread later

[vocal sleet]

@tawny nacelle Has your question been resolved?

[versed bane]

YIPPEEE

[versed bane]

@tawny nacelle what’s the actual question

like is it this?

[vocal sleet]

@tawny nacelle Has your question been resolved?

Channel closed

Closed by @tawny nacelle

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>.
• 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.

[edgy gulch]

is this correct

[vocal sleet]

@edgy gulch Has your question been resolved?

[edgy gulch]

.close

[vocal sleet]

Channel closed

Closed by @edgy gulch

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[rich timber]

Still unable to get used to how exactly u-sub works and how to make a decision on what to set u too

[lone ocean]

U set the variable 'u' to that part whose differential is already sitting in the expression

[rich timber]

so u=pi/x^17

?

[vocal sleet]

@rich timber Has your question been resolved?

Channel closed

Closed by @rich timber

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[real vale]

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

[kind light]

let Aftab's age and his daughters age be some variable, like x and y

try to make the first statement, then the second statement

[real vale]

thanks

[vocal sleet]

@real vale Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[manic lance]

why is this function correct for constant x0, k, j

[vocal sleet]

@manic lance Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[silk comet]

for $j \leq k$, you get
[((x-x_0)^k)^{(j)} = (k \cdot (k-1)\cdots (k-(j+1)))(x-x_0)^{k-j}]
and evaluating at $x = x_0$ gives $0$.

for $j = k$, the above formula gives $k!$, so taking any further derivatives will give 0.

[twin meteor]

Tushar

[vocal sleet]

✅

[eternal inlet]

hi

[vocal sleet]

@manic lance Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[misty owl]

how many sheets of metal that are 500mm x 1000mm can fit in there and can you show calculations

[misty owl]

the little pipe doesnt matter

[vocal sleet]

@misty owl Has your question been resolved?

[misty owl]

<@&286206848099549185>

[vast shale]

So can u send the question again?

[misty owl]

how many sheets of metal that have measurements of 500mm x 1000mm can fit with this image

[vast shale]

What is that don’t break thing ?

[misty owl]

just a sign

its not a part of the actual calculations

[vast shale]

okay

so can u find the area of the image first by using the lengths that are given in the image??

u need help with that?

[misty owl]

im pretty sure its 12.3 x 3 = 37.8m^2

[vast shale]

very good

now what is the area of the sheet

like one metal sheet

[misty owl]

500mm x 1000mm

but i think i should convert to metres

so then they are all the same

[vast shale]

do u know how to convert

[misty owl]

divide by 1000

[vast shale]

from metres to mm

very good

so now tell what is the area of the image in mm^2?

[misty owl]

first convert them

[vast shale]

convert what?

[misty owl]

12300mm x 3000mm = 36,900,000mm^2

[misty owl]

the sides

[vast shale]

very good and what will u get after converting it?

[misty owl]

12300 mm x 3000mm

[vast shale]

very good

[misty owl]

= 36,900,000mm^2

[vast shale]

so now, do u know to calculate sheets now?

basically u have the area of the image,right, now ujust need to divide it by the area of sheet, then it will give me thenumber of sheets

[misty owl]

so 36,900,000mm^2 / 500,000mm^2?

[vast shale]

yes

what would that be?

[misty owl]

73.8

[vast shale]

yes

[misty owl]

omg

thank you so much

[vast shale]

that is correct

its fine

[misty owl]

.close

[vocal sleet]

Channel closed

Closed by @misty owl

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>.
• 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.

[lethal bison]

hi

[lethal bison]

there is a icosceles trapezium ABCD

AB = a, CD = b , a>b

AB is perpendicular to CD

the diagonals AC and BD meet at point O

if angle BAD is alpha and angle BOC is beta

prove that

tan(beta/2) = a-b / a+b tanalpha

i would really apreciate some help

Im onto nothing

<@&286206848099549185>

hey sorry for being annyoign but is anyone there? please

[warm ivy]

In process of solving

[vocal sleet]

@lethal bison Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lethal bison]

.reopen

[vocal sleet]

✅

[lethal bison]

okay brother imw aiting here

[ripe jay]

??

[lethal bison]

oh its a troll

@regal parrot@nimble elm@near smelt

he is trolling

[nimble elm]

this is a mathematics server.

[near smelt]

new user, tried to ping everyone

low effort bait

[lethal bison]

he was member since jan

[near smelt]

That was the first time he'd posted lol

[vast shale]

back to the question

[vocal sleet]

@lethal bison Has your question been resolved?

[ripe jay]

my bad it should be

oh nvm it's obvious i got it all

u wanna hear my solution

or just some hints

oh you guys have 2 russian names so I got confused XD

[ripe jay]

I saw that you worked out the right side

now we need to work out the beta/2

180 - beta may help you

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[velvet cove]

how to do

[velvet cove]

,rotate

[velvet cove]

i got t=8 but that doesn’t sound right

[muted anchor]

,w minimize sqrt((t/2-1)^2 + (t/2-3)^2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by intoxicatedhippo

[muted anchor]

I think that's what you're looking for, you've just gone wrong with the working somewhere

[vocal sleet]

@velvet cove Has your question been resolved?

[muted anchor]

You might as well drop the sqrt to make things a bit easier for yourself too

[lethal bison]

i dont get it pls explain

i mean i undertand what u said but idk how i can implement it

[ripe jay]

let's open another channel this one is occupied

[vocal sleet]

Channel closed

Closed due to timeout

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>.
• 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.

[earnest charm]

How do I solve what I marked?

[strong grove]

let u = 2+r might work

[earnest charm]

and what i need to do with that?

u=2+r

r= u-2

[strong grove]

u = 2+r
r = u-2
$$\int \sqrt u (u-2) du$$

[twin meteor]

JustToPro

[strong grove]

u can multiply sqrtu with both terms and then power rule

(dont forget to change bounds unless if u are going to convert back into r)

[earnest charm]

how u change that to du
u need to multiply dr with u' to get du

[strong grove]

u = 2+r
du/dr = d/dr(2+r)
du/dr = 0 + 1
du/dr = 1
du = dr

[earnest charm]

du/dr = d/dr(2+r)
i didn't understand this line

[strong grove]

differentiate both sides with respect to r

[vocal sleet]

@earnest charm Has your question been resolved?

Channel closed

Closed by @earnest charm

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[vast shale]

simplyfy one

<@&286206848099549185>

[spice ether]

Hey, so I’m working with GL(n,F) in an intro course for group theory. How does one go about proving closure and associative of matrix multiplication in this group? Any advice?

[vast shale]

i am in eight💀

[tidal dock]

#❓how-to-get-help

,rccw

[tidal dock]

which one

oh i see nvm

[vast shale]

can u help ?

[tidal dock]

notice that $b^{1/2} = \sqrt{b}$

[vast shale]

my eight standard brain is bursting

[twin meteor]

artemetra

[vast shale]

wait can u explain me how it is possible

[tidal dock]

it's the same thing

just different notation

[vast shale]

hmmmm ok

[tidal dock]

in general $b^{1/n} = \sqrt[n]{b}$

[twin meteor]

artemetra

[vast shale]

that means a is also the same right?

[tidal dock]

yep

[vast shale]

hmmmmm

[tidal dock]

do you know how to proceed?

[vast shale]

are u good in area and perimeter?

[vast shale]

yes'

[tidal dock]

great

[tidal dock]

i think so

just show the problem

[vast shale]

a swimming pool is surrounded by a concrete path 4m wide if the area of the concrete path surrounded by the pool is 11/25 that of the pool find the radius of the pool

[tidal dock]

okay so

it is helpful to sketch a diagram

[vast shale]

ok

soo ummmm now ?

[tidal dock]

should look something like this

let the radius of the pool be r

[vast shale]

hmmmm

[tidal dock]

thus the radius of the pool and the path is (r+4) meters

does that make sense?

[vast shale]

yes

[tidal dock]

great

[vast shale]

and thatis 11/25

[tidal dock]

no

that's the area

we are getting to that

[vast shale]

yes that only

[tidal dock]

what's the formula for the area of a circle?

[vast shale]

227/7 multiplied by r square

[tidal dock]

okay i'll write pi for simplicity

[vast shale]

ok

[tidal dock]

we have that $\pi r^2 =$ area of the pool

[twin meteor]

artemetra

[tidal dock]

and $\pi(r+4)^2 =$ area of the pool and path

[twin meteor]

artemetra

[vast shale]

hmm

[tidal dock]

thus, $\pi(r+4)^2 - \pi r^2$ is the area of the path

[twin meteor]

artemetra

[tidal dock]

does that make sense?

[vast shale]

hmmm yes

[tidal dock]

so

[vast shale]

we can short out pie

[tidal dock]

since it's 11/25 of the pool

$\frac{\pi(r+4)^2 - \pi r^2}{\pi r^2} = \frac{11}{25}$

[twin meteor]

artemetra

[tidal dock]

that's your equation where you need to solve for r

[tidal dock]

indeed

[vast shale]

but why divide ?

[tidal dock]

because

the area of the concrete path surrounded by the pool is 11/25 that of the pool

[vast shale]

ok'

[tidal dock]

it simplifies to $\frac{(r+4)^2 - r^2}{r^2} = \frac{11}{25}$

[twin meteor]

artemetra

[tidal dock]

can you take it from here?

[vast shale]

yes

[tidal dock]

awesome

[vast shale]

Thank u very much

[tidal dock]

no problem

[tidal dock]

to close this channel, type ".close"

[vast shale]

.close

[vocal sleet]

Channel closed

Closed by @fair oasis

Use .reopen if this was a mistake.

[vocal sleet]

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>.
• 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.

[digital acorn]

Ex 14 compare the 2 numbers

[fallow igloo]

You could put everything in 2 underroot and then compare

[digital acorn]

can you show me an image or write how would it look? If u may ?

[fallow igloo]

I am not home rn but wait a sec I think there somethin on google

[fallow igloo]

I'll do for first one

But I won't write the first under root

Wait

I didn't see the + sign

[digital acorn]

No problems

[fallow igloo]

How do you need to compare the 2 num

Can you square on both numbers?

[tidal dock]

,rccw

[twin meteor]

Couldn't find an attached image in the last 10 messages.

Aug 23, 2024, 2:48 PM

[digital acorn]

Like if one is bigger or smaller than another or if they are equal

[tidal dock]

,rccw

[fallow igloo]

You can square both sides

Get rid on one square root

[digital acorn]

And then ?

[fallow igloo]

Then just put value of ✓6 and✓7

[digital acorn]

Is it possible without aproximating ?

[fallow igloo]

Wait let me think

You can subtract 6 from both sides

Then u get 5+4✓6 and 5✓7

From there you know 1st num is bigger

[digital acorn]

Yeah, ok Thank You Very Much )

[fallow igloo]

Yw

[digital acorn]

.close

[vocal sleet]

Channel closed

Closed by @digital acorn

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>.
• 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.

[vast shale]

okay guys i know its like third time i ask for nearly the same thing, but im close to fully getting it again. tho i like to ask how to figure this one our, the 1, in the middle is kinda confusing me

[vast shale]

Rewrite as an irreducible fraction.)

[daring fractal]

what grade are you in

[twin horizon]

Simplify the top

[fallow igloo]

First write division in another term

[fallow igloo]

Wouldn't it be simpler if you rewrite division as another term first?

[vast shale]

bro i forgot how to fix these tupe of questions

[twin horizon]

Either way works 🙂

[fallow igloo]

Ig that's true

[vast shale]

so 3/12

[daring fractal]

real

[fallow igloo]

Now write division as another term

And then simplify that

[vast shale]

as another term??

[twin horizon]

So that's ((3/12)/(1/(3/8)))

[fallow igloo]

@twin horizon
Can u explain i gtg

[vast shale]

aaah, okay

[twin horizon]

Have you learned indices

Like powers and exponents

@vast shale

[vast shale]

ehm

[twin horizon]

That's not the final answer

[vast shale]

english iss

idk

[vast shale]

yeah so

like this?? then wwhat?

or wait, u want me to write it sideways??

[twin horizon]

It turns to (3/12)÷(3/8)^-1

[vast shale]

okay

let me type it

[twin horizon]

what would x^-1 become

ping when you are ready

[vast shale]

wait

si

wait that aint how to write it

where to place the 1

@twin horizon

[twin horizon]

Do you understand this

[vast shale]

x multiplied with itself -1 times

??

[twin horizon]

Yes but express it as a fraction

[vast shale]

eehmm

eeeeeeeeeeeeeeee

???????????????????+

[twin horizon]

x^-something=1/(x^something)

[vast shale]

so how do i type it?

[vast shale]

like

where put it

the 1

[twin horizon]

It seems like you haven't learned indices

I will explain another way

[vast shale]

yeah perhabs.