help-49

[heavy lynx]

sorry so is module here the r?

in the euler form

[distant heron]

yes

[heavy lynx]

is that root x^2 +y^2

or wait

uh

[distant heron]

exactly

[heavy lynx]

so 1 = root x^2 +y^2

or

OG

OH

hold on so

if i do

ln(re^i theta)

then the real part is ln(r) and the imaginatry is (e^i theta)?

omg wait is this like the log rule where if you multiply something inside the log u add it outside

then the i comes out

OHH

ok its coming together lol

tysm

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[heavy lynx]

could someone please explain this

[midnight plank]

Please don't occupy multiple help channels.

[heavy lynx]

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[mystic salmon]

Let n(x) = x^2 - 4. The average rate of change of n(x) over the interval [c, 5] is equal to 3, where c is a constant. Find the value of c.

my work:

(n(5)-n(c))/(5-c) = 3
(21-n(c))/(5-c) = 3
21-n(c) = 3(5-c)
21-n(c) = 15 - 3c
-n(c) = -6-3c
n(c) = 3c + 6

[midnight plank]

@mystic salmon Has your question been resolved?

[mystic salmon]

<@&286206848099549185>

[brazen palm]

you can use n(x) to find n(c) in terms of c and then solve the quadratic to find c

[mystic salmon]

yeah i tried that while i was waiting

and it simplified to

c^2 -3c-10 = 0

which i solved and got c=5, c=-2

can't be 5 bc other side of interval is 5

so has to be -2

but i don't get how it can be -2

[brazen palm]

oh... lemme check

I'll try it my way to see if i can get c=-2 too

[mystic salmon]

oh nevermind i got it

when i was trying to see if the slope between (-2, 0) and (5, 21) worked, i was using a slope of 1/3 instead of 3

ok ty for clarifying

[brazen palm]

cheers!

[mystic salmon]

have a good rest of your day

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[brazen palm]

you too

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[tardy patrol]

Hi isnt this supposed to be 4sinxcosx?

Cause there was already a 2

So sin2x is 2sinxcosx

[inland oriole]

yes

[tardy patrol]

Therefore 2.2sinxcosx?

aigh thanks

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[timber tartan]

"Which is the odd one out"

[modest forge]

all of them seem to be polygons except a certain shape...?

[timber tartan]

B???

[modest forge]

yea

[timber tartan]

Oh wait what

Had to google what a polygon is rq

Thank u

[modern sapphire]

Well, google regular polygon and you get C as odd one

[modest forge]

circle isn't a polygon though?

[modern sapphire]

it is as n goes to infinity

[robust forge]

I must say What a profound question of math

[timber tartan]

This question was ffom our school exam

U take the exam to get accepted to the school

[modest forge]

dude that looks like a elementary school question though

[modern sapphire]

fair

[timber tartan]

It is

[robust forge]

Nvm u dont get the sarcasm

[timber tartan]

Its for 6-7th graders

[timber tartan]

People going to 7th

[modern sapphire]

But, still circle, triangle and square are kinda regular figures

unlike C

[modest forge]

yea I can see that too

[last slate]

A group theorist might say the answer is C

[modest forge]

just realized that this sounds aggressive, i didn't mean that lmfao

[timber tartan]

Bro we havent learnt ab infinitt

😭

Infinity

[robust forge]

[timber tartan]

Is the answer b??

[modern sapphire]

(cries in passing uni but failing 6th grade)

[unreal fjord]

I mean, it can just be A for not properly aligning with the another answer. I kinda hate this kind of question.

[robust forge]

Should be c since rest of the shapes are symmetrical

Or rather -c

[timber tartan]

My friend said d cuz all other shapes are 360 degrees 😭😭😭😭😭

[modest forge]

the person who designed the question might as well have ocd and consider C as the answer for not being neat-looking

[timber tartan]

LOL

[timber tartan]

😭

[robust forge]

I mean that ain't wrong either

[timber tartan]

Yeah so it might b A for not having a reason 🥶

[robust forge]

was thinking that only

[timber tartan]

Which is it now

😭😭😭

[modest forge]

your safest option is still probably b though

[timber tartan]

More u think ab the question more complicated it gets

Annoying

[robust forge]

Case 1 should be circle since it has infinite sides
Case 2 should be trapizium since it's not symmetrical
Case 3 should be triangle cause it's the only figure with angle sum 180
Case 4 should be triangle because it is placed a little above (misprint)
Case 5 should be A because it's the only one not being noted down in any of the 4 cases

[timber tartan]

What is a trapizium? And i actually didnt know circles had infinite sides we didnt leadn that 😭

[robust forge]

it must be a millennium problem

[modest forge]

(because most likely they want you to be thinking about geometry, and polygons are, well, a huge chunk of geometry)

[robust forge]

Trapizium is quadrilateral with atleast 2 of the sides being parallel

[robust forge]

Case 6 should be circle since it's the only implicit curve here

[modest forge]

what language is this by the way?

[timber tartan]

Kurdish

[timber tartan]

😭😭

[timber tartan]

Oh

Thanks

[timber tartan]

Letters and writing looks really similar to arabif

arabic

they have like 20 of the same letters

I think

[modest forge]

yeah I figured it's kurdish but was surprised they were allowing the language on exams

didn't mean to get political on a math server though

[timber tartan]

Wdym?

wdym allowing the language on exams

😭

[modest forge]

sent u a dm, if anyone else is reading this sorry that this got off topic

[timber tartan]

Thanks for the help everyone

[robust forge]

gonna report this and then u will get banned and then u won't be able to get math help ul fail uni and then ul go into depression and then commit suicide

[robust forge]

What's the final answer

[modest forge]

my career is over

[timber tartan]

Idk i wouldve chosen B

[timber tartan]

LOL

[robust forge]

Indeed it is

[robust forge]

U mean -B

[timber tartan]

ima be using this server alot starting in sep

[timber tartan]

LMAO

its js like that cuz like our writing is the opposite way of english's writing

english left to right ours is right to left

they usually put abcd in our language too idk why they did it in english this time

[robust forge]

ohh

[midnight plank]

@timber tartan Has your question been resolved?

[midnight plank]

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[timber arrow]

im rlly bad at these types of qs where i gotta use trial and error

cuz i dont rlly know where to actually start

[shell wigeon]

Well if you use the smallest denomination, 3, to get to 8, you also need 5, and if you add another 3 you can get to 11 as well.
Now to cover the rest (9,10,11), you only need an additional 4.

(unless it's a trick question and you just buy 9+3 to get 12 which covers 12 and everything under)

[timber arrow]

ohh so 3+5+3+4 so 15

[timber arrow]

was thinking this asw

but it means exact values

[timber arrow]

this is such a simple quick method, starting from the lowest and working to eac hvalue thanks

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[shell wigeon]

The question also kinda proves it works because it certainly cannot be 12 (any combination that only gets to 12 can't also get to all the other ones), and the next choice is 15, which works

[shell wigeon]

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[real pivot]

Other channel closed due to inactivity so Im asking this again

[real pivot]

How did they guess T(n) = O(n) here

I asked other ppl from my class and they arent sure either

Idek what the recursion tree would look like for something like this

[visual tiger]

saying floor(n/2)+1 is approximately n/2, using the recursion you would have T(n) = n + n/2 + n/4 + ...

[real pivot]

so I should draw the recursion tree for T(n/2) + n?

[visual tiger]

i would just take n = 2^k - 1

and write T(n) in terms of k

[real pivot]

how did you get this

[real pivot]

also if I do this then the work done each layer would be just n right?

or would it be 1

Im getting O(n log n)

coz we have log_2(n) layers and work done on each layer is n

so n log_2(n) = O(n log n)

[visual tiger]

it's not n on each layer

T(n/k) is not n + ...

[real pivot]

I dont get it

[real pivot]

what do u mean by this

[visual tiger]

you have log_2(n) layers sure

but the value of T on each layer changes

it's not n+n+...+n log_2(n) times

[real pivot]

but arent we making one recursive call each time?

so I thought the work done each layer would be the same too

[visual tiger]

?

say we start with T(n)

[real pivot]

right

[visual tiger]

then we write it as T(n/2) + n (oversimplifying the recursion in T)

then T(n/4) + n/2 + n

then T(n/8) + n/4 + n/2 + n

each recursive call you make, you make it on a smaller and smaller number

[real pivot]

Oh so we have
T(n/2^k) + sum_(i = 0)^(k-1) n/(2)^i

wait no i mean

[visual tiger]

up to k-1 but yes

[real pivot]

so now I pick k = log_2(n)

[visual tiger]

yes, tho this isn't the exact same relation

I approximated floor(n/2) + 1 as n/2

[real pivot]

oh ye

[visual tiger]

though it's either n/2 + 1 or (n+1)/2

if you want the best first case to look at

[real pivot]

so what would I do after this

[visual tiger]

i suggest n = 2^k - 1

[visual tiger]

n + n/2 + n/4 + n/8 + ...

what does this look like

[real pivot]

idk

approximately n?

oh geometric series?

n/(1-1/2) = n/(1/2) = 2n

and now I can ignore the 2 so I get O(n) ?

[visual tiger]

yeah

<= 2n (limit of the infinite sum)

and >= n (first term)

n <= T(n) <= 2n

[real pivot]

so if we had floor(n/2) + 100000

it would still be the same

right

the constant inside wont make a diff

[visual tiger]

wdym

if the recursion is T(n) = T(n/2) + constant

now each layer adds the same "computation time"

[real pivot]

nah

T(n) = T(floor(n/2) + 1000000) + n

[visual tiger]

For a large enough initialisation, i would say yes probably

[real pivot]

Also one more question. This was from my midterm and this is how I was doing it. I dont understand the issue with T(floor(n/4) + 1) + n/2 + n here

am I not expanding it in a similiar way as u did

[visual tiger]

well all of the remarks are in there

as I said I was oversimplifying T(n) recursion

I wrote it as T(n/2) + n instead of T(floor(n/2)+1) + n to avoid some problems

your question was how to get the intuition

[real pivot]

but thats allowed right? i just wanna be sure incase something like this pops up on my exam tmr

[visual tiger]

if you wanna prove it formally, show it for some form of n that actually consistently simplifies floor(n/2)+1

here you gotta use n = 2^k -1

that's what I would do

[real pivot]

or could I just do all of that somewhere else to get the correct guess and instead just do what my prof did here

coz you dont have to show how you got the guess when doing questions like these right

[visual tiger]

yeah there is nothing wrong with tweaking the question a bit to get the correct intuition

but for formal proofs you gotta be as rigorous as can be

[real pivot]

Im gonna be honest I still dont understand where exactly Im suppose to plug this n + how you got n = 2^k - 1

[visual tiger]

well the teacher uses a different method of "we have the guess, so prove it by strong induction"

[real pivot]

that should be fine too right

[midnight plank]

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[drifting root]

help with this pls

[midnight plank]

@drifting root Has your question been resolved?

[drifting root]

<@&286206848099549185>

[pastel spoke]

Just my thought

You could apply the law of cosine

And get two expression with cos(theta) and r

Then you should be able to find the r by solving them

[drifting root]

?

ohhh

i see your vision

[pastel spoke]

Sure

[pastel spoke]

Do you have the answer

If so, I might give it a shot to do the calculation

[drifting root]

ans is 20

[pastel spoke]

For sure

Let me do a quick calculation to confirm it

[drifting root]

ye it works

[drifting root]

thanks mate

[pastel spoke]

No problem

[drifting root]

i think ill wait to see what the dude is typing tho lmao

[fringe haven]

Call angle PAB by θ.

Cosines rule on PAB gives us-
4h²=16h²+90²-8h•90cos(θ)
Cosines rule on APC-
9h²=16h²+120²-8h•120cos(θ)

These are two equations with variables cos(θ) and h.
From the first one we get-
cos(θ)=(3/2)h+90

Substituting into the second equation we get-
9h²=16h²+120²-8h•120((3/2)h+90)
Which is a quadratic you can solve for

[pastel spoke]

He seems to got a big brain plan

[drifting root]

i just set for cos instead of subbing in

dealing with expanding is a pain lol

thanks for the help

.close

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[twilit field]

question c

[twilit field]

I started with the coefficent matrix

$\begin{bmatrix}
1 & 1& -1&2&1\
1 & 2 & -1&1&1\
2&3&-1&2&1\
4&5&-2&5&2\
\end{bmatrix}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

Then $R_3-2R_1$

[grand pond]

f(why am i here )= I don't know

[twilit field]

$\begin{bmatrix}
1 & 1& -1&2&1\
1 & 2 & -1&1&1\
0&1&1&0&-1\
4&5&-2&5&2\
\end{bmatrix}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

After a few more operations, I arrived at

$\begin{bmatrix}
1&1&1&1&1\
0&1&1&-1&0\
0&0&0&1&-1\
0&1&-6&1&-2\
\end{bmatrix}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

But this doesn' really help me all that much

because of $a_{42}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

I think I got it

Thanks chat!

.close

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[night egret]

i guess bro just needed a space to work

[twilit field]

I didn't think of something

I wrote down all my steps

but didn't think of interchanging a few columns

[subtle blaze]

we dont normally do that

we only exchange rows..

[twilit field]

My bad

used the wrong word

I meant rows

[midnight plank]

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[twilit field]

Find the reflection of $e^x$ about $x=2$

[grand pond]

f(why am i here )= I don't know

[twilit field]

so this basically means f(x)=g(4-x) intutively

But how would I solve this more rigourously

Like I'm not sure how to explain myself

but what I know is that the graph should be a reflection so it will be $e^{-(x +a)}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

and as it has to be 1 , 4 units from O, a=-4

so the equation would be $e^{-(x-4)}$

[grand pond]

f(why am i here )= I don't know

[sullen elm]

Hi guys

Can someone help

[twilit field]

,w graph e^x and e^{-(x-4)} from 0 to 4

[grand pond]

Results provided by WolframAlpha

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

Requested by physicsrocks

[twilit field]

#❓how-to-get-help

Like just a hint please

Basically, I have$e^x$

[grand pond]

f(why am i here )= I don't know

[twilit field]

and a plane mirror at x=2

I have to find the equation giving it's reflection ( Of the virstual image)

It is.

But that was based off pure intution.

I was hoping for a more conceptual approach

Again, just a hint, please

Wait, do you have an alt on this server.

👍

OK, let me think a bit more then

OOh

I can think of this in this manner.

Now I know that $e^x$'s reflection is $e^{-x}$

[grand pond]

f(why am i here )= I don't know

[twilit field]

now I want the reflection about x=2

So, I shift the origin

Now, I want e^{-(2-a)}=e^2$ at x=2

that gives me a=4

Thanks !

.coose

[plain osprey]

i think u messed it up bud

[twilit field]

I have another question

Find the reflection of e^x about y=4

-e^x +a =e^x

so a=4

oh

right

I thought it was $-e^x - 8$ for some reason

[grand pond]

f(why am i here )= I don't know

[twilit field]

,w graph $y=-e^x and y=-e^x+8 from 0 to ln(4)

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[twilit field]

.close

[midnight plank]

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[waxen silo]

this is physics related but this part of it is more maths how do you figure out the proportionality of two different things? I never really got the hang of it

[waxen silo]

related formulas

you don't need to know what they mean but since star 2 has 2x the lifetime of star 1

how does that effect it's proportionality?

is it really 1/2*(star1)^3?

[shell wigeon]

T proportional to 1/M^3 just means T = k * 1/M^3 for some k

Solve for M

[waxen silo]

what does the k stand for here

[shell wigeon]

A number

[waxen silo]

ah, so since the lifetime is 2T

I assume move 2 to the other side

so it's 1/(2*M^3)

[shell wigeon]

...no

T = k * 1/M^3

Solve for M

[waxen silo]

oh make m the subject

M=cuberoot(k/T)

right?

but what is k in this specific situation

[shell wigeon]

An unknown constant

You can call it "constant of proportionality" if you wish, I don't care

[waxen silo]

ah

[shell wigeon]

We don't know its value because it's not given, all thats given is a proportionality relation

Anyway

[shell wigeon]

Yes

Now these M and T are for the first star

We want M_2, and we know T_2 = 2T

We know it's the same k

[waxen silo]

yes

so it cancels out when we sub it in?

[shell wigeon]

So same formula

M_2=cuberoot(k/(2T))

Now since you don't know T, you need to use M (or rather M_1) instead

[waxen silo]

u huh...

[shell wigeon]

What would be M_2/M_1?

[waxen silo]

cuberoot(k/(2T))/cuberoot(k/T)

[shell wigeon]

Right

Can you simplify that?

[waxen silo]

I got cuberoot(T_1/(2*T_2))

are the T's the same in this situation?

[shell wigeon]

Yes that's not T_2

[waxen silo]

because if so it's just cuberoot(1/2)

ah

[shell wigeon]

Yes

Which is...?

[waxen silo]

0.7937....

guess that's my answer

or do we still need to relate it to M_1

[shell wigeon]

That's M_2/M_1

[waxen silo]

ah I see, move M1 to the other side

I got 1.786*10^30

[shell wigeon]

,calc 2.250 * 0.7937

[grand pond]

Result:

1.785825

[shell wigeon]

Yeah, I guess it's 4 significant digits

The question says "approximately" anyway...

[waxen silo]

yea lol

how did you know to put Mass 2 over mass 1 btw?

why not mass 1 over mass 2?

[shell wigeon]

That doesn't matter

[waxen silo]

or equate them via equations

[waxen silo]

oh

would you get the same thing either way

[shell wigeon]

If you compute M_1/M_2 you get 1/0.7937

Just solve for M_2

[waxen silo]

ah

makes sense

I see now, thanks for the help

[shell wigeon]

[waxen silo]

going to quickly review this then close

[shell wigeon]

If you want to check your results, a question to ask yourself is whether it makes sense that M_2 is less than M_1, and also that it's more than half of M_1

[waxen silo]

do u know the physics behind it?

alr

.close

[midnight plank]

Channel closed

Closed by @waxen silo

Use .reopen if this was a mistake.

[waxen silo]

.reopen

[midnight plank]

✅

[shell wigeon]

You don't have to, it just comes from the given proportionality relation

[waxen silo]

oh

yea didn't think of that, thank you

.close

[midnight plank]

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[frigid turret]

can someone explain me what is a unison sign

[midnight plank]

Please don't occupy multiple help channels.

[frigid turret]

how does it look like i mean

[robust forge]

U mean union

[hollow coral]

$\cup$

[grand pond]

BoiledAnchovies

[frigid turret]

and interceptioni sign

or whatever u call it

[frigid turret]

the opposite of this

like the n

[robust forge]

Intersection

[hollow coral]

yes

[viral dagger]

$\cap$

[grand pond]

Skill_Issue

[frigid turret]

Ahh ok

got it

ty

.close

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[vague seal]

what have you tried?

[last slate]

this is fun problem

[drifting root]

i tried saying smth about the midpoints in each side

[last slate]

good

start

[drifting root]

but like

you cant really say the diagonals intersect at where the midpoint lines intersect right

so im stuck

[last slate]

try writing in terms of vectors what u said

[vague seal]

you have some vectors u and v

and you have the diagonals u+v and u-v

[drifting root]

yep

[vague seal]

in order to show that the diagonals bisect eachother

is to show that u+v/2

and u-v/2 in that diagram are the same location

but since u-v/2 does not start from the origin

aka A

you need to add something to it to make it start from the origin

[drifting root]

uhhh idk what this add smth is, but can i just say DC=AB in a parallelogram and for the other side too

[drifting root]

also, how does this work

[vague seal]

to bisect is to say that it splits it in two

which means that the intersection of the two has to happen at u+v/2

and u-v/2

if u-v/2 were to start from A

[drifting root]

i thought

it would be at 1/2(u+v)

[vague seal]

yes

[drifting root]

or am i getting smth terribly mistaken

[vague seal]

sorry

my bad

[drifting root]

o

[vague seal]

(u+v)/2

:D

[drifting root]

lol

[vague seal]

me lazy

:p

[drifting root]

nwnw

you can type it like that from now on

[drifting root]

i get that part now

i just have a hard time actually proving it

[vague seal]

You want to show that (u-v)/2 if it were to start from A instead of where it starts now is the same vector as (u+v)/2

In other words

the intersection of the diagonals

is where some multiple of the diagonals is equal

apart from the fact that again u-v does not start from the origin A

and so needs to be translated to the correct starting point

first

[drifting root]

ok wait a sec

lemme

absorb

if i say that those are equal, arent i saying that AP = DP?

[vague seal]

no

that would be saying the lengths are equal

what you want to say is that the vectors are the same

you want to say

if i go from the origin

along u+v

for some time

i will come across u-v

at some multiple of u+v

a(u+v)

and if i go along u-v

etc etc

v + b(u-v)

that make any sense?

[drifting root]

this is gonna take a while and lot of repetitive q from me sorry

so

for no reason at all

i thought |u+v|=|u-v|

but theyre vectors

so theyre not equal

but if this were a rectangle or smth, then it would be equal?

and this is why i got it mixed up with (u+v)/2 and |(u-v)/2|

[vague seal]

a square

[drifting root]

mk

thats one down

so

[drifting root]

this thing

if im travelling down u+v, and suddenly half way i come across u-v where its 1/2 way, its bisecting

[vague seal]

yes

but the idea is to show that it is halway

halfway

[drifting root]

but i still dont understand why they have to be equal

could i get an eg

[vague seal]

because in vector terms

the intersection is the same location

[drifting root]

oh youre right

[drifting root]

so back to this

it doesnt start from a, so how can we still say theyre equal

[vague seal]

look back at the drawing you made

you start at A

and you want to go along u-v

towards the intersection\

without lifting your finger/pen or whatever

[drifting root]

woundt that make it go off the drawing?

like

down

[vague seal]

what do you mean

[drifting root]

im so aorry

ok wait

uh nvm

[drifting root]

oh

huh?

i though BD was u-v

wound that make going along DB v-u?

[vague seal]

Take u

and add -v to it

and draw a line from A to that point

the direction of that vector is from D to B

[drifting root]

oh shit i wrote this wrong

whoops

okok

so working back from what we wanna prove

once we say that theyre equal cause AP, AP, and we say theyre 1/2 each other, theyre bsiecting

this sisnt even working backwards idk why i said that

[drifting root]

so how do we start proving this now?

[vague seal]

a(u+v)

v + b(u-v)

by showing that when these are equal a and b are both 1/2

a(u+v) is the red line

and v + b(u-v) is the orange line

[drifting root]

i forgot where the v+ came from sorry

[vague seal]

in order to start travelling along u-v

you first have to travel along the entirely of v

so to get to P

[drifting root]

ah

[vague seal]

you go along v

• some amount of u-v

[drifting root]

icic

[vague seal]

so v + b(u-v)

[drifting root]

ok

so do i just make an eq where they are equal?

to start off

[vague seal]

yes

that's the idea

when they are equal it means you're at the intersection of u+v and u-v

[drifting root]

?

why is this legal

it feels illegal

[vague seal]

why does it feel illegal?

wait

how did you go from a + b - 1 = 0 to a+a = 1

did you do the left first?

[drifting root]

yep

sorry should have said sub it in or smth

[drifting root]

it feels weird to just say u(…)= 0 and v(…) = 0

[vague seal]

you should remember that you can only say ku + lv = 0 => k = 0 & l = 0 if u isn't a multiple of v

[drifting root]

cause theyre not unit vectors or perpendicular

[drifting root]

yea

[vague seal]

it is not illegal

it is a property of linearly independent vectors

actually a defining property

[drifting root]

i learnt a good chunk of vectors by myself cause i missed my lessons sprry

[drifting root]

how can we tell if a vector is linearly independent

[vague seal]

a set of vectors $\left{ v_1, ... , v_r \right}$ is linearly independent if and only if the equation $a_1 v_1 + ... + a_r v_r = 0$ has only the solution $a_1 = 0, ... , a_r = 0$

[grand pond]

Katharine

[vague seal]

a vector itself doesn't have the property of linear dependence or independence

a set of 2 or more do

in this case u and v

are a set of 2 vectors

and since there is no other solution to

$k v + l u = 0$

[grand pond]

Katharine

[vague seal]

then $k = 0$ and $l = 0$

they are linearly independent

[grand pond]

Katharine

[drifting root]

ok

i have some other vector q’s

could i get confirmation on this one

ill send a pic

[vague seal]

why the dots?

also lazy? XD

[drifting root]

yep

lol

[vague seal]

i think it works

[drifting root]

also

[vague seal]

guessing that you've defined the length squared of a vector as the dot product

between itself

right

[drifting root]

yes

[vague seal]

|a + b|^2 = (a + b) . (a + b)

[drifting root]

let me write one out

uh

(a+b).(a+b)

=a.a + a.b + a.b + b.b

and the abs val

[vague seal]

i know the dot product is commutative

but if you're explicitly writing both terms

try to remember that the second one is b.a

[drifting root]

ic

[vague seal]

in this case it doesn't matter

for 2 reasons

but in other things it mighjt

anyway

[drifting root]

is there a case that you can show where it might matter?

[vague seal]

matrices

[drifting root]

ah

nvm

idk what that is lol

[vague seal]

(A + B)(A+B) = AA + AB + BA + BB

and in matrix multiplication

it isn't always commutative

meaning AB is not always BA

[drifting root]

ic

also uhhh im really sorry to be bothering you with the same type of question but theres another one

[vague seal]

this is a matrix

[drifting root]

PH THIS THISNG

[vague seal]

PH?

[drifting root]

oh

[vague seal]

ah

[drifting root]

i have typing prolems

ah

prfoblem

problems

[vague seal]

multiplying two of these

two sums of these

as in (A+B)(A+B)

gives AA + AB + BA + BB and AB is not BA

sometimes it might be

but generally no

anyway

[drifting root]

wait wait uhhh

let me try

[vague seal]

did you try drawing this?

[drifting root]

from here

i forget how to relate b back

alsp

i wrote this wrong

there

man

i wrote this

really wromg

[drifting root]

@vague seal is this one alright?

[vague seal]

that seems good

[drifting root]

alr

one more q to go

is a because

i find AP

and AC will be a constant*AP

its smaller than AP so its between 0 and 1

for b tho, how would i prove concurrent

mmmmmm

find OR first?

[vague seal]

you show that the intersections of AP and BQ and AP and OR have the same point

that would show the lines are concurrent

you get BQ and OR for free

so don't try to prove all three combinations

[drifting root]

?

[vague seal]

If you prove that the intersections of AP and BQ as well as AP and OR are the same point

then you don't have to prove that the intersection of BQ and OR also is the same point

and vice verse

[drifting root]

oh iv

[vague seal]

proving BQ-OR and AP-BQ have the same point means you don't have to prove that AP-OR also has the same point

[gaunt imp]

Can't he use the proposition that the segments that go from one vertice of a parallelogram to the opposite one intersect in the middle point as a lemma?

I mean, that should save him quite a bit of work, and it's a pretty well known property of parallelograms

[vague seal]

question 12 is a bout a triangle

can you explain

[gaunt imp]

Oh, nvm then, for some reason thought it was a parallelogram

My bad, haha

Same thing still works for the orthocenter in a triangle though, if I'm not mistaken

[drifting root]

ok

so this

[drifting root]

yea but they want a vector method

im stuck on showing the intersections have the same point

i can find the ratio

i think-

i hope this is the right way

@vague seal this is what i have, but it doesnt feel correct enough

what am i missing?

[vague seal]

if you know lambda 1 then this would prove the 2:1 ratio

for OR

[drifting root]

lambda 1 = lambda 3

how does it prove OC is concurrent tho

[vague seal]

it doesn't

it proves that the ratio of OR to OC and AP to AC is the same

In order to prove concurrence

you need to show that OR-AP intersect at C

and also

OR-BQ intersect at C

lets call the intersection of OR and AP C

that's a 'given'

[drifting root]

yep

[vague seal]

then you need to prove concurrence by showing that the intersection of OR and BQ is also at C

that would show that three lines are intersecting at 1 point

which is concurrence

[drifting root]

since the q gave that BQ and AP intersect at C, are we allowed to do this?

[vague seal]

use that one

the intersection of BQ and AP is C

then prove that the intersection of OR and AP is also C

ignore what i said about C before

[drifting root]

ok

uhh

can we go though it together

and ill try for the other one

[vague seal]

The goal is to prove that the equation you used in you ratio thing is valid

that you can in fact say AC = AO + OC vector wise

that OC is a thing

[drifting root]

oh

so i just

sub 2/3 in my OC eq, add AO and if its eq then concurrent?

[vague seal]

Lets call that equation OD

to not confuse ourselves by using something that implies we know for a fact it's concurrent

[drifting root]

ook

[vague seal]

so if you can show that OD + AO = AC then you have proven concurrence

as then you have proven that D and C are the same point

vector addition

btw

for OD + AO = AC

[drifting root]

so

lambda 3 OR

=OD

[vague seal]

yes

[drifting root]

ahhh i got it now

[vague seal]

D is the intersection between AP and OR

and that happens at some lambda_3 multiple of OR

[drifting root]

yep

tysm for the last like

2 hrs

[vague seal]

i was playing a game too :D

[drifting root]

o lol

have fun playing

.close

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[fallen heath]

lets take a topological space X. Is it possible for $A , B \subseteq X $ , $G_\delta$ sets to have that $ A \cap B = \emptyset$ but $\overline{A} \cap \overline{B} \neq \emptyset$ ?

[grand pond]

whitefang

[fallen heath]

actually yeah it can like in R, A= (0,1) and B=(1,2)

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[fallen heath]

lets say we have an infinity topological space X and an equivalence relation on X with [x] being an equivalnt class for $x \in X$. what requirements we need for the equivalence relation (or the space X) so that this implication would be always true for $ x,y \in X$ if $\overline{[x]} = \overline{[y]}$ then $[x]=[y] $

[grand pond]

whitefang

[last slate]

By $\overline{[x]}$ you mean the closure of $[x]$?

[grand pond]

Bequi

[last slate]

@fallen heath

[midnight plank]

@fallen heath Has your question been resolved?

[fallen heath]

Yeah the closure

If we have that there is a equivalent class which is dense on X then it fails

But no dense equivalent classes is enough to ensure it?

[last slate]

Not necessarily. Consider if you had the equivalence classes {0} and R\{0} in R. One of them is dense but your condition still holds here

And no dense equivalent classes also won't be enough

[fallen heath]

Oh actually yeah you are right

[last slate]

You could have an equivalence relation containing the classes (0,1) \cap Q and (0,1) \cap Q^c in R

Neither of those are dense in R

but their closures are still equal

This is kinda tautological but what you need is that for every closed set X there is at most 1 equivalence class which has X as closure

[fallen heath]

Sorry right now I am not in a position to check if I understand it but I will check it later

Thank you

[midnight plank]

@fallen heath Has your question been resolved?

[midnight plank]

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[inland patio]

Consider the metric $d(x,y)=\max{d_1(x_1,y_1),\ldots,d_n(x_n,y_n)}$. Obviously, $d_k(x_k,y_k)\leq d(x,y)$ for all $k$, hence convergence in the product space implies convergence in the component space. But how can I show convergence in the component space implies the convergence in the product space? I know how to do this for the Euclidean metric, but here I am stuck. I feel like I need some upper bound to $d(x,y)$.

[grand pond]

psie

[nova yoke]

you need the following basic observation:
d(x,y) < epsilon if and only if d_k(x,y) < epsilon for all k

[inland patio]

ok, that was helpful, thanks 👍

.close

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[thorny oasis]

why?>??

[thorny oasis]

i understand everything before it

why is current through A and B 0.10 A and curernt through C 0.20 A????

[hard shard]

The current is split so that voltage is equal on either side

[thorny oasis]

I don't get it

[hard shard]

the equivalent resistance of the top line is 120 ohm

by using V=IR, we know that the top line has twice the resistance and therefore must have half the current

[thorny oasis]

oh okay that makes sense

but why is D = 0.3?

is there a way to show it mathematically

or like through an equation

[hard shard]

do you mean I?

[thorny oasis]

and wont C = A and B?

[thorny oasis]

I?

[hard shard]

oh D

[thorny oasis]

current across like indivisual lod or whatever it is

[hard shard]

are you familiar with kirchhoffs laws?

[thorny oasis]

umm no

[hard shard]

eh thats fine

[thorny oasis]

like i thought, at D there was 60 resistance right? so
60/100 * 0.3

[dull yoke]

Me when garlic garlicks

[hard shard]

bruh

[thorny oasis]

that doesnt make sense

IM SO CONFUSED

like how did they get their values

0.1, 0.2, 0.3, for the indivisual lamps

mathematically, is there some sort of process or steps i can follow?

[hard shard]

the current flowing through a circuit must stay constant unless it splits at a junction

[thorny oasis]

right

[hard shard]

or if two currents converge at a junction

you can imagine the first three lightbulbs as a single resistor with 40 ohm resistance

which makes this easier to think about

[thorny oasis]

right okay

so 2 resistors in series,
first is 40, second is 60

[hard shard]

because then you have one loop with no junctions so the current must be constant throughout

[thorny oasis]

right, makes sense

[hard shard]

electricity problems is a lot of fiddling around with things until they make sense to you

[thorny oasis]

ive got an exam on all of circuits tomorrow

20 more pages in textbook

and i cant understand the part which was supposed to be a given

🦾 🦾 🦾 🦾

[hard shard]

keep practicing and youll get the hang of it

[thorny oasis]

ITS 3 AM 😩 😩 😩

[hard shard]

oh my

best of luck to you

[thorny oasis]

why thank you

have a great day yourself

.close

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[cunning loom]

i'm not sure exactly why i'm getting this wrong

[inner belfry]

is this SAT stuff?

[cunning loom]

no 😭

data and financial lit

i'm a 12th grader 💀

[inner belfry]

Can you go through how you did this

[inner belfry]

Nice

[cunning loom]

ok for the first answer i just multiple the mean by day by the # of days

so

247 * 10

[inner belfry]

ugh hold on

okay yeah

go on

Is it refering to total distribution or distriution of food and lodging

[cunning loom]

Daily standard deviation: $60 ----> Number of days: 10
multiplying the daily standard deviation thing by the square root of the number of days the standard deviation of the overall cost over a period of days.

[cunning loom]

that's what i'm trying to figure out 😭

[inner belfry]

this is above my math level sorry 😭

[cunning loom]

i swear the answer should just be ^2 b (which should be 360k)

[cunning loom]

no worries i might just drop this question bc this is so aids 😭

[inner belfry]

never taking stat

[midnight plank]

@cunning loom Has your question been resolved?

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[naive sleet]

could someone

[naive sleet]

Explain to me keypoint 2.11 in detail

I dont understand why they are parallel

[midnight plank]

@naive sleet Has your question been resolved?

[naive sleet]

<@&286206848099549185>

[woeful escarp]

wsp

[night egret]

multiplying f(x) by a constant only scales the output of the function, not the input

since the output is y, it only scales the function in the y-direction

which is the same direction as the y-axis and hence parallel

[naive sleet]

can you

make me understand

my showing me this

by drawing a grap

thing is

i understand it

but i cant visualize it

[gusty birch]

,w graph sin(x)

[grand pond]

Results provided by WolframAlpha

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[gusty birch]

,w graph 10sin(x)

[grand pond]

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[gusty birch]

,w graph sin(x) and 10sin(x)

[grand pond]

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Requested by woomygfx

[gusty birch]

heres one example

you can play with it (in #bots )

[naive sleet]

but why is it "parallel"

[gusty birch]

cause its getting stretch only vertically

[naive sleet]

so its getting stretched "relatively" from the x-axis but parallely from the y-axis

THIS IS MY ISSUE I CAN VISUALIZE IT BEING STRETCHED RELATIVE TO X-AXIS

BUT THE PARALLEL STATEMENT DOESNT MAKE SENSE TO ME

[gusty birch]

yea im trying to also figure out how to explain the parallel part

[naive sleet]

thank you <3

[gusty birch]

ah i have an idea

this is the graph for sin(x)

[naive sleet]

yes

[gusty birch]

lets draw the x and y value for sin(pi/2)

so

[naive sleet]

yes

[gusty birch]

now lets stretch sin(x) by a factor of 10

[naive sleet]

okie

[gusty birch]

notice how the line parallel to the x axis stays the same lenght

but the line parallel to the y axis get stretched

thats what it means by being "stretched parallel to the y axis"

is it clear or do you need me to explain further?

[naive sleet]

but

bro

oh you said line paralle to axis

okays

[gusty birch]

yea

[naive sleet]

makes sense to me

thank you'

i will give you cookie if i ever met you

[gusty birch]

ahahaha thanks

[naive sleet]

goodbye

.close

[midnight plank]

Channel closed

Closed by @naive sleet

Use .reopen if this was a mistake.

[gusty birch]

my fav is obviously chocolate chip (maybe with some vanilla)

[naive sleet]

send aaaadddddddddddreeeeeeeeees

[gusty birch]

xd

[midnight plank]

Available help channel!

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[fallen heath]

does basic nbhd means nbhd basis ?

[midnight plank]

@fallen heath Has your question been resolved?

[wet reef]

Hello everyone, this is video i made about why does the area under the curve of 1/1 + x**2 equals PI, the proof is visual and uses concepts like probability and geometry to respond the question https://www.youtube.com/watch?v=V2gv-BP9SpU

[fallen heath]

sorry probably you posted it in wrong channel

[midnight plank]

@fallen heath Has your question been resolved?

[midnight plank]

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.

[twilit field]

Express the function $\frac{1}{\sqrt{x+\sqrt{x}}}$ As the composition of 3 functions

[grand pond]

f(why am i here )= I don't know

[twilit field]

I was thinking $f(g(h)))= \frac{1}{\sqrt{x+\sqrt{x}}}$

[grand pond]

f(why am i here )= I don't know

[proud island]

ye

[twilit field]

f(x) =1/x

[wary trail]

i was about to say 1/2sqrt(x) lmao

[twilit field]

$g(x)= \sqrt{x+\sqrt{x}}$