help-36

[ivory hazel]

I don’t know how to format it properly though

I know the triangles are AAS

[frail moat]

Do it

[fossil kiln]

As in how to write it...?

[odd seal]

proving congruency at this point is just redundant tbh

[ivory hazel]

Something like that

I got this answer from an answer key but I’m not really understanding it still

[fossil kiln]

Angle t would be equal to angle u

Do yk why

[leaden moon]

What do you have so far now? First, what does the question give you first?

[ivory hazel]

The question gives me the given and the prove so I know how to start off

But that’s about it

[leaden moon]

And can you try to see why two angle S equal to each other?

[ivory hazel]

Yeah

[leaden moon]

How about the 2 angle S?

These 2 angles are the key to this question

[leaden moon]

@ivory hazel

[ivory hazel]

Uhh

[fossil kiln]

Those triangles r already given congruent why do u need tk prove....?

[leaden moon]

I don't see that mentioned in the question?

[fossil kiln]

Oh wait thts the angle not the triangle lol

[leaden moon]

😆

[ivory hazel]

They’re on opposite sides so it’s vertical angles right

[leaden moon]

Vertically opposite angles yes

[ivory hazel]

Alright

How do I write the proof

[fossil kiln]

There r multiple ways

You could prove the triangles congruent or use sum of angles

The given would be angle T =U

And the vertically opp. Angles

Then NS=SH

[ivory hazel]

So N = H

[fossil kiln]

This would imply the triangles r congruent by AAS

And hence the asked angles would also be equal

Do y get it....?

[ivory hazel]

Yeah a little

[fossil kiln]

What r u unsure abt

[ivory hazel]

Is it ok if we do a couple of more problems

[fossil kiln]

Mmhm

[ivory hazel]

alright give me a second

I don’t really feel confident in this one

[ivory hazel]

@fossil kiln

[fossil kiln]

SAS

Do yk tht?

@ivory hazel

[ivory hazel]

Side angle side

[fossil kiln]

Yup

So we do have 2 sides given equal?

[ivory hazel]

I think so?

[fossil kiln]

Its in the ques

[ivory hazel]

oh💔

[fossil kiln]

XD

Now what do we need?

To apply SAS

[ivory hazel]

Uhhhh

[fossil kiln]

We need 3 things

We've got 2, which are the the pairs of equal sides

[ivory hazel]

I don’t know I’m so confuse

[fossil kiln]

Okay the criterian is SIDE ANGLE SIDE

[ivory hazel]

Ye

[fossil kiln]

Those r 3 things tht need to be equal for triangles to be congruent

U understand this?

[ivory hazel]

A little

How do we write the proof though

[fossil kiln]

You just write 3thingd tht r equal

And why

The pair of sides are given equal

We need the the angle between them to be equal now

Ie B

[ivory hazel]

Wait

[ivory hazel]

To each other or?

[fossil kiln]

Each other ofc

[ivory hazel]

Ohhhh

So SAS and ABD congruent EBC

[fossil kiln]

Yass

[ivory hazel]

So one more then

Triangle ABD congruent to EBC?

[fossil kiln]

Yeah

[ivory hazel]

ohhhhhhh

So we just proved it now

Can we do just one more?

Or nah

[fossil kiln]

Ur choixe sry

Tag me pls

[ivory hazel]

Alright

[final saddle]

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[ivory hazel]

Alright thank you

[final saddle]

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[rustic wedge]

there aer 12 points in a plane 5 of which are collinear. find the max numbeer of distinct quadrilateral formed with vertices from these points

[rustic wedge]

i did complementary counting and got one of the right answer, which is 420

but there was 1 more correct option, and i wanna know if theres a more direct way to get it

namely 2*7P3

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[fossil kiln]

Total- not possible?

[rustic wedge]

yes

[fossil kiln]

You'd need to make cases

@rustic wedge

[frail moat]

Wdym 1 more correct option

[fossil kiln]

2 × 7p3 i think

[serene panther]

I think hes talking abt another method

To get the same answer

[frail moat]

K

[rustic wedge]

yes

[fossil kiln]

What abt cases?thts a longer method tho

[serene panther]

I think its better if we focus on 2*7P3 since thats what op wants

[frail moat]

where is op anyways

[fossil kiln]

Mmhm but i hv no idea where tht comes from

[fossil kiln]

🥀

I remember asking this ques to chatgpt

And it accounted for 3 collinear and 1 not too lmao

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[gritty monolith]

can anyone help me with this

[gritty monolith]

not solve it but any clue could be appreciated

b number

[scarlet sequoia]

do you know the time at which the the tube hits the ground

[gritty monolith]

no i tried finding it

but i dont think we can

[muted prairie]

<@&268886789983436800>

[zealous storm]

yikes

[gritty monolith]

...

[candid pulsar]

high risk operation

[gritty monolith]

we dont know the intial velocity as well

[muted prairie]

It's knocked off the tower

[scarlet sequoia]

you know how many feet the tube drops in t seconds

[muted prairie]

You're only worried about vertical velocity

[scarlet sequoia]

and you know the height of the tower

[gritty monolith]

i tried implementing s = ut + 1/2 at ^2

is this correct approach

[muted prairie]

Is this physics or calculus

[gritty monolith]

i found acceleration its -16m/s^2

[gritty monolith]

calculus, but whenever do these kind of question, i think of physics

I mean physcis

[scarlet sequoia]

you're given that h(t) = C - 16t^2

C = initial height

[muted prairie]

then this isn't the intended approach

[muted prairie]

this i mean

[gritty monolith]

hmm

so to calculate the last 2 seconds

first we have to find the time taken

[frail moat]

if its calculus then knock doesn't imply u = 0

[gritty monolith]

otherwise its no way

[gritty monolith]

okay..!

[scarlet sequoia]

well, sure

but you have h(t) = 400 - 16t^2

and you want the time it takes to reach the ground

[gritty monolith]

hmm

let me think

ok we can use derivative i guess? and that would give us velocity? and we know velocity is 0 when it reaches the ground

ah..

[scarlet sequoia]

Velocity right before you reach the ground is really high

there is no moment when you reach velocity = 0

because velocity is discontinuous

What is special about the ground

where is it?

[gritty monolith]

ah..

i get it

[shrewd turtle]

Anyone know who can i darw it ?

[gritty monolith]

400-16 t ^2 = 0

[scarlet sequoia]

!occupied

[final saddle]

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

[gritty monolith]

so t would be 5 seconds!!

[shrewd turtle]

Sorry this

[scarlet sequoia]

again, occupied

[shrewd turtle]

I don’t know what he want to draw in b

[gritty monolith]

so now,

the question becomes average speed in the first 3 secocnds, which is ,

16 times 3 ^ 2 / 2

[scarlet sequoia]

This isn't your channel. It's occupied by @gritty monolith. If you want help with your question, open a channel

Also you already have a channel that belongs to you

#help-21｜아리스킨충1 @shrewd turtle

[shrewd turtle]

Okay thanks u

Soo much

[scarlet sequoia]

what

[gritty monolith]

i meant

the question asks us last 2 seconds

and we have formula

spped = distance/ time

distance would become 16 * t * t / 3

since the question is asking us about "average speed", we use the abvoe formula, i suppose?

[scarlet sequoia]

average speed over a time span = (distance traveled)/(time taken)

[gritty monolith]

yes!

[scarlet sequoia]

I agree the time taken is 2 seconds

but what is the distance traveled??

it's the distance between the initial position and final position

[gritty monolith]

the distance traveled is 16 * t ^2 , as we know from the question, the test tube drops at 16 t^2 feet

[scarlet sequoia]

which are...

[scarlet sequoia]

from the start

not from any point

do you think that in the final second, the test tube drops 16 feet?

[gritty monolith]

oh waittt

400 - 16 t ^ 2

/3

[scarlet sequoia]

/3?

[gritty monolith]

yes

?

[scarlet sequoia]

take time to answer this

what is the initial position in "the final two seconds"

what is the final position

[gritty monolith]

final position is 400 and inital is 16 * 3 * 3

so
difference between them is 2 seconds ?

sorry my bad

[scarlet sequoia]

you're mixing so many things up xdd

[gritty monolith]

yes

its 2 seconds

not three

[scarlet sequoia]

If you wanna switch from the usual height to "position from the top"

then it's gonna change the formulas but Ig we'll end up with the correct result if we do things correctly

but why not just use height

[gritty monolith]

final position is 0 and intial is 16 * 3 * 3

[scarlet sequoia]

height(initial) = height after 3 seconds

[scarlet sequoia]

h(3) = ?

[gritty monolith]

h(3) = 400 - 16 * 3 * 3

h(5) = 400 - 16 * 5 * 5 = 0

h(5) - h(3) / 2

i think

thanks raphaelisius

for helping me out!!

the answer is 0 - (400 - 16 * 3^2) /2 = 128 ft/sec

[scarlet sequoia]

the answer is technically negative first but it's because it's velocity

speed is the absolute value of that

[gritty monolith]

yes

[scarlet sequoia]

and so negative bc goes down

[gritty monolith]

Thank you.

.close

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[mossy rampart]

which part is cauchy schwarz ineqality? Not sure what exactly it is.

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[proper dagger]

top one looks like Jensen, though not sure if it is

[mossy rampart]

Oh yeah that's holder's

Thanks though!

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[devout vale]

help me integrate this, i tried 3 times still dont get it

[chilly haven]

what did you get stuck on?

[devout vale]

after this step what to do im confused

[vital crag]

if you have the whole solution, just show it all

[devout vale]

yes i have but im not able to understand how it moved further wait 1 min

[gusty goblet]

Hello

[gusty goblet]

Guys, I am pretty confused with this question as my Sir had reduced marks for no reason as this is quite an easy one so it got me thinking what i got wrong in this

A customer wants to buy 3 books from a book shop. 2 books are non fictioned and their MP is 580, the other is a fiction book marked at 750. If the shop offers 20% discount on all items, then what is the total bill amount the customer would have to pay?

[full jungle]

tbh you could substitute tanx itself as some variable squared

[gusty goblet]

i got bill amount as 1528 after discounted all the things and the total

[chilly haven]

this ones taken bro get ur own channel

[gusty goblet]

ohh mb

im new

[keen vector]

#❓how-to-get-help

[devout vale]

cos?

[full jungle]

its easier to work with normal variables instead of trigonometric functions

[devout vale]

ok ill try

wait

[vital crag]

they used a trick called king's rule

[keen vector]

Applying kings i think makes sense but not sure it they want to ise that here

[keen vector]

I think so too

[full jungle]

@devout vale oh yeah use kings rule instead, substituting will make it long i forgot it exists

[keen vector]

Its usefullness has no bounds

[full jungle]

agreed.. until queen steps in

[keen vector]

We dont talk about queens

[full jungle]

bruh

[vital crag]

#discussion or #serious-discussion

[dusty helm]

try kings rule by replacing x with pi/2-x

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[vital kelp]

i'm trying to prove every normal p group is contained in a maximal normal p group i've proven that every normal p group belongs to the sylow p group P. How do i start ?

[vital kelp]

lebesque integral > riemann integral

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[vital kelp]

<@&286206848099549185>

<@&286206848099549185>

oops

my bad for double ping

[worldly mesa]

zorn?

I mean you just have finite set of normal p groups ordered by inclusion, so of course it has a maximal element

[vital kelp]

doesn't it require proving the chain has maximum element and all but because it's finite, this is trivial

[worldly mesa]

yeah yeah

I explained more

[vital kelp]

interesting I actually got a proof from Gemini which is neat I guess this works as well

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[heavy grove]

Hello, I'm trying to learn Sturm-Liouville theory, but I have no idea what's going on. Can anyone point me to some good resources or maybe give me the run down?

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[rustic torrent]

wtf

[hybrid heath]

Try one of the advanced channels, or wherever your topic applies

[heavy grove]

alr thx

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[vital kelp]

isn't nil trivially nilpotent ?

[abstract mango]

by nil you mean the zero matrix?

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[vital kelp]

no i mean nil ideal

but i got my answer

thanks

and thank you gemini

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[dreamy moon]

Can someone help please

[vital kelp]

which question ?

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[shut relic]

Was wondering on how to convert infix to postfix for ternary operation. I'm using shunting yards algorithm and it works well for binary and unary operators but for tenary I'm having trouble on handling ? : notations and how to represent them.

Would a ? b : c (infix) be a b ? c in postfix?

[tardy quartz]

no, the postfix for the ternary infix expression $a \text{ ? } b \text{ : } c$ is not $a \ b \ \text{?} \ c$

[soft zealot]

alsx

[shut relic]

would it be a b ? c : then? I'm a bit confused on how they are converted

[tardy quartz]

do you know The fundamental rule of Reverse Polish Notation (RPN), or postfix, is?

[shut relic]

not formally but we would have our operator follow our operands like so a b +

and we would evaluate it left to right

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[shut relic]

does it not change when converting? e.g,

infix a > b ? c : d
postfix a b > ? c : d

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[shut relic]

.close

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[plain rain]

,tex
hello there. so our prof gave us the following exercise:
\ Let $a_n , n\in\mathbb{N} $ be a sequence satisfying:
\ \ $ a_{2n} , a_{2n-1} \to l , \ l\in\mathbb{R} \ , n\to\infty $ . Prove that $a_n \to l , \ n\to\infty $ . My thinking is that i should approach the problem by using the fact that each one of those two convergent sequences have all their subsequences converge to l as well, but i have no idea how to continue. In fact, perhaps characterising the two sequences' outputs as members of sets would be a better way. Sorry if i yapped too much. Please guide me towards the right approach, be it one of mine or something else

[soft zealot]

fijokazż

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[plain rain]

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[timber plume]

question c, this is what I have tried

[tired walrus]

do you have the definition of quartile on hand

cause it sounds like you completely forgor

& instead decided to do something that smells vaguely 25%-ish

[timber plume]

it divides into 4 sections

q2 is the median

q1 is the lower 25 percent

q3 is the lower 75 percent

[tired walrus]

mmmmm dodgy wording.

Q3 is the answer to the question "75% of your sample lies below which point?"

or, in the case of a RV, 75% of the area under the curve

in other words, Q3 is the value at which the CDF equals 0.75.

if your calculator is capable of calculating the inverse CDF for a Gaussian,

then Q1 and Q3 are the values of that at 0.25 and 0.75 resp.

[tired walrus]

note that these are NOT equal to the median value ± 25% of the median value! there is NO such percentage-based relationship that can be derived at all between them!

[timber plume]

whats inverse CDF?

[tired walrus]

do you know what CDF is

[timber plume]

and based on my understanding of question 3, I think we need to calculate the Q3, Q1 and IQR

[timber plume]

nope

[tired walrus]

then you could have like

asked me what that meant

CDF stands for Cumulative Distribution Function

does that sound at all familiar to you

[timber plume]

I never heard of it

i heard of cumulative frequency

[bold turtle]

-# In fairness, you're the one who brought CDF up as a term; it's not in the question itself

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[jagged flare]

how do you prove something is "oppositely oriented"?

[jagged flare]

i got the first bit of 4.35, proving the triangles are simmilar

[whole halo]

this means one triangle is a mirrored version of the other

its enough to use that they share the same side MI, but MK and MT are on the same line

the hard part is knowing how to write this rigorously

[jagged flare]

can you use the fact that like MK>MI, and MT>MI?

(actually is this even true)

[steady locust]

where is this from, just curious

[jagged flare]

euclidian geometry in mathematical olympiads by evan chen

[steady locust]

oh that makes sense

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[jagged flare]

.solved

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[waxen otter]

What is Fourier series?

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

[pliant shore]

https://www.youtube.com/watch?v=r6sGWTCMz2k

[waxen otter]

Thanks

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[sharp sandal]

how would i solve question 4 b

[sharp sandal]

.close

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[jagged flare]

4.39

[jagged flare]

the hint was ATM_c=STM_b, then angle chase

but where do you angle chase this?? am i blind

like the only thing i can reasonably think is proving
BTS=CTS
BTM_c+ATM_c+ATS=STM_b+M_bTC
ATS=(b-c)/2 (we want to prove this)

but i legit dont know how to prove that

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

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@jagged flare Has your question been resolved?

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[lone fog]

i dont get why answer isnt

C

like

isnt it always of the form

(1+a)^(a non natural power)

[vital crag]

(a + b)^r is more general

[final saddle]

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[radiant carbon]

not sure how to do this

i know dV/dt = 0.4

V(cone) = (pi(r)^2 * h)/3

r = 16h/20

so V(cone) = 256(pi)(h^3) / 1200

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[silk fjord]

Now I suspect there are two parts to this problem:

1. find the area
2. make it 3d

i think apothem can be found easily as 4

like this

then $A=\frac{1}{2} \cdot 4 \cdot p$

[soft zealot]

UCYT5040

[silk fjord]

idk how to find perimeter here though

then also

for step 2

area of circle is $\pi r^2$

[soft zealot]

UCYT5040

[silk fjord]

volume of a sphere is $\frac{4}{3} \pi r^2$

[soft zealot]

UCYT5040

[silk fjord]

iirc the volume of a sphere definition is complex w/ integrals

but however it is found

i suspect a similar operation is to be performed

on the area of the hexagon in this problem

anyone have any thoughts on this?

[severe verge]

not sure why you want the volume of a sphere for this

[silk fjord]

i dont

im saying

the way area -> volume with circle -> sphere

must be similar to this problem

[summer snow]

I believe you've made a slight mistake when identifying AD; it may help to label the points of the hexagon.

[severe verge]

here's our regular hexagon and AD is our rotation axis

[silk fjord]

oh yeah thats true

[severe verge]

now you should be able to imagine what kind of shape it makes after the rotation

[silk fjord]

is it a cyllinder with two cones

[severe verge]

exactly

next time, label your points carefully

[silk fjord]

but wait

how do i find the radius?

its half BF or CE

[severe verge]

radius of the bases of the cones and cylinder you mean?

[silk fjord]

yes

[severe verge]

what kind of triangle is this

[silk fjord]

right

so pythagorean can be applied but

i dont have AG or AF

[severe verge]

remember that our hexagon is a regular hexagon

[silk fjord]

oh maybe its a special right tri let me check

[severe verge]

so you should be able to get the angles of that triangle

[silk fjord]

its 30-60-90 right triangle

[severe verge]

awesome

now we need to find AF

we just use the idea that we can divide a regular hexagon into 6 equilateral triangles

this should solve all your problems

[silk fjord]

so side length is 4?

[severe verge]

yep

[silk fjord]

can you do that with any regular polygon?

or just hex

[severe verge]

only a hexagon

[silk fjord]

ok

[severe verge]

because the angles in a regular hexagon is 120°

so the angles in each triangle would be 60°

this isnt the case for other regular polygons

[silk fjord]

oh yeah that makes sense

anyways i find radius of $2\sqrt{3}$

[soft zealot]

UCYT5040

[silk fjord]

height of cylinder is 4 (8-2-2)

and cone height is 2

now just plug into formulas

i got $72+16\sqrt{3}$

[soft zealot]

UCYT5040

[silk fjord]

for k

can someone check the answer key for me?

http://www.white-oak.net/HS/Questions/Reg/R07A.pdf

[severe verge]

wait that sounds wrong

[silk fjord]

this is question 9 on the geometry test

oh

oh no

wait

i used surface area

not volume

for cone

😭

one sec lemme redo

ok

i got

64

can you check?

i dont want to look at the key myself it will ruin all the other questions

[severe verge]

64 is correct

[silk fjord]

cool!

thank you so much for your help!

.close

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[winter raft]

Prove that the unit group $U_n$ where $n$ is an odd composite number cannot be cyclic.

[soft zealot]

Allicin (Alli)

[winter raft]

uhh i dont know where to start

wait i got the problem wrong

its just asking about the product of two distinct primes

well i know that totient(n)=(p-1)(q-1) if n=pq

n is still odd though

so that means p and q are odd

and p-1 and q-1 are even

im not sure how that helps

[final saddle]

@winter raft Has your question been resolved?

[winter raft]

abstract algebra is throwing me for a loop

ok well

if the group is cyclic

powers of an element must generate the whole group

lets say the element is g

well

uh

[plucky rover]

Well g^|U_n| = 1

And no earlier power is 1

You should be able to take it from here

[winter raft]

i cant find an earlier power that makes a contradiction

[plucky rover]

You sure?

[winter raft]

uh

what?

[plucky rover]

I meant

You can use that to find an earlier power

Bad phrasing sorry

[winter raft]

i feel like im missing something obvious

[plucky rover]

Or maybe I am treating this as easier than it is

Let me work it out and see

Okay while working it out my first idea is CRT

Have you done that

[winter raft]

CRT?

[plucky rover]

Chinese Remainder Theorem

[winter raft]

oh right

let me look that up really quick

ohh

we can break it into mod p and mod q

[plucky rover]

It is easier than I thought as well, but not the way I thought

[winter raft]

we need g^n congruent to 1 mod p and q

[plucky rover]

And then you can use that the direct product of cyclic groups is cyclic if the orders are coprime

[plucky rover]

Have you done this

[winter raft]

i dont think so

[plucky rover]

Ah

[plucky rover]

Well you can still get there using this

Good luck

I got an exam cya

[winter raft]

we do have g^n congruent to 1 mod p

thank you

we can think about Up

great movie

i liked Kevin

ok anyway

Up has order p-1

so n|p-1

we can say the same for Uq to get n|q-1

wait

no

what

hold on

g^(p-1) is congruent to 1 mod p

g^(q-1) is congruent to 1 mod q

p-1 is even and q-1 is even

i feel like thats important

wait yeah it is

g^(k*(p-1)) is congruent to 1 mod p

hold on ill brb

ok i got it

.close

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[final saddle]

@granite tapir Has your question been resolved?

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[timber plume]

Im tryna sketch the graph

[timber plume]

I have all the info required but I am finding some contradictions

if x approaches infinty, f(x) approaches 0

so why is there an x intercept?

[rugged merlin]

The function is not monotonic

I.e., its not always increasing

[timber plume]

for e), I may have done the it wrong, I just plugged in a value of 100, then a big value such as 10000000 and saw the difference

[timber plume]

its decreasing

[primal sparrow]

You could just plug in infinity

[timber plume]

you cant

in the calc

[primal sparrow]

And l’hôpital

[timber plume]

?

[rugged merlin]

Not for all of its domain

[bronze grove]

you only have to draw the curve right?

[timber plume]

yes

[bronze grove]

take that property on derivative

not the function

[timber plume]

wdym?

[bronze grove]

firstly do u have the derivative of the function?

[timber plume]

yes

[bronze grove]

now can u tell increasing or decreasing according to the derivative?

hint : look for when f'(x) = 0

[timber plume]

ill come back to it its a bit tricky

[versed cargo]

Is f(x) = lnx/x?

[final saddle]

@timber plume Has your question been resolved?

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[jagged flare]

4.39

[jagged flare]

the hint was ATM_c=STM_b, then angle chase

but where do you angle chase this?? am i blind

like the only thing i can reasonably think is proving
BTS=CTS
BTM_c+ATM_c+ATS=STM_b+M_bTC
ATS=(b-c)/2 (we want to prove this)
but i have no idea how to??

[raven marsh]

egmo

[jagged flare]

ikr

[raven marsh]

TA symmedian of triangle TKL

TI is the median

@jagged flare does that help?

angle chase or smth

[jagged flare]

TA is symmedian and TI is median is used for 4.38

angle chase is already in the hint

[raven marsh]

😭

[jagged flare]

idk what you expect from me man

[raven marsh]

what

im as cooked as you are

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

[robust mulch]

sorry skish it isnt for lack of trying, im just bad at constructive geo

[final saddle]

@jagged flare Has your question been resolved?

[tranquil pine]

Doesn't this give you ATMc = STMb

[jagged flare]

yeah?

[jagged flare]

this

[tranquil pine]

wait are you asking for the angle chase that comes afterwards

?

If yes then idk why you were trying to use T in the angle chase

[final saddle]

@jagged flare Has your question been resolved?

[jagged flare]

yeah?

[jagged flare]

what else then?

[tranquil pine]

if you know ATMc=STMb, S is not defined in terms of T anymore

Say you only have ABC and Mc, Mb, can you identify S from just these points?

it's more of a heuristic thing but you should always step back and see if you can reduce the problem to an easier one like here that you don't need T anymore

Basically the angle condition gives you ||ASMcMb is an isosceles trapezoid|| so you know all the angles and you can angle chase easily

[jagged flare]

ohh wait

ok i see, thank you!

.solved

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[stone cradle]

how do i make the normal form of an layer E like this with the given points A(1,1,-3) , B(0,2,2) , C(2,1,-5)

[stone cradle]

hope it makes sense because it was in german and i tried to translate it

[severe hawk]

!original

[final saddle]

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

[stone cradle]

8 a

[mental roost]

Using A B C you can get n

AB and AC are vectors in the layer
Then AB x AC will be perpendicular to it

[stone cradle]

n is the vectorproduct right ?

[mental roost]

n is the normalized vectorproduct of two vectors in your layer, yes

Once you have that you can sub one point in the equatio and solve for a

[stone cradle]

is n (-2,7,-1)?

and is this the solution ?

wait i forgot an - on the 3

[mental roost]

The easiest check is to input the layer into desmos/geogebra whatever and then also graph the points

Or input the points to the equation and check if it is 0

[stone cradle]

what do i do in vector x then ?

to check if its ß

0

[mental roost]

Put in point A

Treat it like a vector

Do that for B and C as well

Should be 0 for all 3

[stone cradle]

oh i got it

lemme check

[final saddle]

@stone cradle Has your question been resolved?

[stone cradle]

yo

on b it is 4

not 0

[mental roost]

Thats not good

Try graphing E A B and C in desmos

[stone cradle]

whats that

never heared about it

[mental roost]

Graphing website

[stone cradle]

that wont happen

no clue how that works

[mental roost]

Im deutschen ist geogebra verbreiteter

[stone cradle]

bist du deutsch?

[final saddle]

Channel closed

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[stone cradle]

.reopen

[final saddle]

✅ Original question: #help-36 message

[stone cradle]

pls help

[final saddle]

@stone cradle Has your question been resolved?

[drowsy epoch]

what is the frage

[stone cradle]

deutsch?

[drowsy epoch]

bro ich seh dich hier jedes mal

[stone cradle]

ich war mir nicht sicher ob du es bist

hab das gerade geschafft zu lösen jetzt habe ich aber ein anderes problem

undzwar von der normalengleichung zur parametergleichung

E4

es gibt ja mehrere lösungen oder?

[drowsy epoch]

normalenform -> koordinatenform -> parameterform

[drowsy epoch]

ja

[stone cradle]

kann man das auch ohne die koordinatenform machen?

die hatten wir noch nicht glaube ich

[drowsy epoch]

ja aber dann ist dass keine richtige von Form A nach Form B kommen

[stone cradle]

hä

[drowsy epoch]

man kann theoretisch, egal welche form, immer zwei vektoren nehmen, die in der ebene liegen, und sie als richtungsvektoren wählen

dann noch einen beliebigen punkt und fertig

[stone cradle]

ich verstehe

aber wie finde ich die richtungsvektoren heraus?

wenn du willst kann ich dir die buchseite dazu schicken

check ich aber nur so halb

[drowsy epoch]

nimm zwei punkte die in der ebene sind, daraus bildest du einen vektor

[stone cradle]

über übung 9 ist ne erklärung die ich versucht hatte anzuwenden hat aber irgendwie nicht ganz geklappt

[drowsy epoch]

Hätte eher gesagt Hinweis als Erklärung

aber es ist immer das gleiche, du wählst drei punkte, einen wählst du als Ortsvektor, und dann bildest du mit dem und den anderen beiden zwei spannvektorne

[stone cradle]

ok

wie bekomme ich nen zweiten punkt in der ebene?

ich habe ja schon den einen (5,2,0)

das andere ist ja kein punkt sondern der normalenvektor n

[drowsy epoch]

achso ja drei punkte erstmal zu finden ist bisschen arbeit

so kann man sich das vorstellen

[stone cradle]

ich habe das versucht was im buch ist und habe halt schonmal 1 richtungsvektor

hab mir halt was ausgedacht damit es 0 ergibt

ist das richtigß

@drowsy epoch 🙏

[drowsy epoch]

was soll das sein

[stone cradle]

das wie im buch

[drowsy epoch]

ah ok

ja geht

stimmt

du kannst ein kreuzprodukt machen

[drowsy epoch]

mit den beiden dann findest du den dritten

[stone cradle]

das ist das problem

wie

[drowsy epoch]

weißt du wie das geht kreuzprodukt

[stone cradle]

ja

[drowsy epoch]

ja dann führ es aus

[stone cradle]

oh ok

ist das richtig?

[stone cradle]

das erste ist mit dem kreuzprodukt und dann die endlösung

[drowsy epoch]

welche war das

E4?

[stone cradle]

ja

[drowsy epoch]

,w (r(1,1,0)+s(0,0,2))*(1,-1,0)

ja

das passt

[soft zealot]

Results provided by WolframAlpha

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

Requested by adoniszgks

[drowsy epoch]

wenn du das (die parameterform) in die normal form für x einsetzt kommt 0 raus

[stone cradle]

hä

kannst du das genauer erklären?

ich verstehe das von wolfram net ganz

[drowsy epoch]

Ich habe nur Wolfram ausrechnen lassen, aber eigentlich sieht man es ja schon

[stone cradle]

oh ich verstehe, nur noch eins. was hast du für r und s eingegeben?

also welche werte?

[drowsy epoch]

nichts

das sind reele parameter

die heben sich weg

das bedeutet also dass die parameterform aufgeht oder anders, egal welcher punkt der parameterform (egal was r und s sind), jeder punkt der parameterform liegt auch auf der normalform

also müssen beide ebenen identisch sein

[stone cradle]

ist zwar unordentlich, sieht aber so aus oder ?

[floral nova]

<@&268886789983436800>

[drowsy epoch]

sieht schön aus

du solltest aber noch r und s hinzufügen

sonst zeigst du nur dass der punkt für r=s=1 das gilt

aber du willst es für alle zeigen

[stone cradle]

oh

also ist es egal ob ich für r und s 100 oder 5 oder ka einsetze es kommt immer 0 raus ?

ist r = s ?

[drowsy epoch]

yep

soll ja auch sein sonst wäre es nicht die gleiche ebene

[drowsy epoch]

nein, nur ich meinte dass beim weglassen technisch gesehen r und s beide 1 sind

[stone cradle]

ja check

nur noch wirklich eine letzte sache

[drowsy epoch]

hau raus

[stone cradle]

ich verstehe jetzt nicht ob ich wenn da mal ist kreuzprodukt machen soll oder skalar

[drowsy epoch]

kreuzprodukt ist mit kreuz also X

das ist ein punkt also ist es immer skalarprodukt

kreuzprodukt mahc schon allein deswegen keinen sinn, weil du rechts 0 erwartest, also eine zahl keinen vektor

ein kreuzprodukt liefert nen vektor keine zahl

[stone cradle]

ok, das hat mich die ganze zeit vorhin auch verwirrt, da die lehrer das bei uns wie nen kleines o hatten also nen punkt mit ner hülle

[stone cradle]

ja

[drowsy epoch]

fürs skalarprodukt?

[stone cradle]

ja

[drowsy epoch]

ja normal

[stone cradle]

wie hier halt

[drowsy epoch]

also wenn ein lehrer fürs kreuzprodukt iwas kreisförmiges nutzt ist der einfach verloren

[stone cradle]

nein für skalar

[drowsy epoch]

ja das ist normal

[stone cradle]

ah perfekt

[drowsy epoch]

meistens will man die "skalarmultiplikation" nicht mit der matrixmultiplikation verwechseln

deswegen unterscheiden da manche

um den kontext klarer zu machen

aber kreuzproudkt ist eig immer sicher ein kreuz

[stone cradle]

Ok, Vielen Dank für die ganze hilfe

war sehr hilfreich

[drowsy epoch]

gerne gerne

[abstract mango]

gut gemacht

[drowsy epoch]

danke danke

[stone cradle]

.close

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[final saddle]

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[elder schooner]

how do i solve the equation 4h(x)=sqrt{2(x+1)}

[scarlet sequoia]

you got a few square roots, how about squaring the equation

[elder schooner]

,, \frac{16x}{x^2 + 2x + 1} = 2(x + 1)

[soft zealot]

admoon

[elder schooner]

so we have this

[scarlet sequoia]

can you try multiplying by the denominator

[craggy bloom]

yes now you would want to get rid of the fraction

[scarlet sequoia]

this way we get a polynomial

easier

[elder schooner]

8x= (x²+2x+1)(x+1)

8x=x^3+3x²+3X+1

wait

,, 8x = \bigl(x + \sqrt{3}\bigr)^2 - 2

[soft zealot]

admoon

[elder schooner]

is it this?

[final tangle]

no

[elder schooner]

wait no i did error

[craggy bloom]

not entirely sure. By guessing your going to need question b

[scarlet sequoia]

you don't need to do that

put everything on one side

[elder schooner]

ohhh okay

[craggy bloom]

From my rough translation question b is telling you one of the roots

[elder schooner]

yeah

,, x^3 + 3x^2 - 5x + 1 = 0

[soft zealot]

admoon

[elder schooner]

we know that 1 is a solution by question b

[scarlet sequoia]

So how do we find the others?

(if they exist)

[craggy bloom]

Is question C to find the positive solutions?

[shell condor]

You can try by depressing the x² term

[scarlet sequoia]

Remember that if a is a root of a polynomial, then you can ||factor by x-a||

[shell condor]

(Or maybe you all are onto something already)

[elder schooner]

guys how do we factor by x-1 i forgot

[shell condor]

Yeah if we already have a root then solving a cubic is fairly simple

[shell condor]

Through long division

[scarlet sequoia]

long division is a way

[elder schooner]

ok wait

[scarlet sequoia]

otherwise, write down x^3 + 3x^2 - 5x + 1 = (x-1)(ax^2+bx+c)

and find the coefficients by developing

[elder schooner]

i find

,, x^3 + 3x^2 - 5x + 1 = (x-1)(x^2 + 4x - 1)

[soft zealot]

admoon

[craggy bloom]

yeah so the other solutions would be found by x^2+4x-1=0

[elder schooner]

yeah i am gonna do this

,, x^3 + 3x^2 - 5x + 1 = (x-1)(x + 2 - \sqrt{5})(x + 2 + \sqrt{5})

[soft zealot]

admoon

[elder schooner]

,, ;so the solutions are ; 1, ; -2 + \sqrt{5}, ; -2 - \sqrt{5}.

is it this?

[soft zealot]

admoon

[elder schooner]

for 3) a- i have just to show that h(x) = f(g(x))

,, f \circ g(x) = \frac{g(x)}{1 + (g(x))^2}

[soft zealot]

admoon

[elder schooner]

,, f \circ g(x) = \frac{\sqrt{x}}{1 + (\sqrt{x})^2} = \frac{\sqrt{x}}{1 + x} = h(x)

[soft zealot]

admoon

[elder schooner]

.close

[final saddle]

Channel closed

Closed by @elder schooner

Use .reopen if this was a mistake.

[lyric obsidian]

,w roots of x^3+3x^2-5x+1

[soft zealot]

Results provided by WolframAlpha

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

Requested by exiled_hype

[lyric obsidian]

you figured everything out? or you still need help?

[final saddle]

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[lyric obsidian]

You can still claim it @elder schooner

[lyric obsidian]

I'll just pin the message

[elder schooner]

i figured it out thanks

[elder schooner]

no you can close the ticket

[lyric obsidian]

okie

.close

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[drowsy epoch]

Find the absolute extrema of f on A

[soft zealot]

Renato

[gentle zephyr]

whatisthegeneralmethodofsomethingthisexercuses

whatisthegeneralmethodofsolvingthis

becausetheremustbeacertainalgorithmtofollow

[drowsy epoch]

iirc check if there are possibly local extrema inside the region, and compare them to the boundary, else they must lie on the boundary necessarily

[gentle zephyr]

canyougointomoredetailonhowtoimplementthatalgoeithm

stepbystephowwouldyousayshouldthisgo

[drowsy epoch]

why are u removing whitespaces

anyway i am off

gl

[gentle zephyr]

isjustfastertotypethisway

[drowsy epoch]

i wonder what u do with your saved time

[gentle zephyr]

ok I will add whitespaces

[gentle zephyr]

the way I usually do this is splitting A into two parametrizations of f

then I find the maximas and minimas for that parametrisation

but what you said about boundary and interior sounds cool aswell, any idea on how to implement that?

[flint sleet]

did you learn what a "gradient" is yet

[gentle zephyr]

Just an object containing the first order partials

vector object or whatever

yeah I heard you can also use lagrange multiplier for this

[final saddle]

@gentle zephyr Has your question been resolved?

Channel closed

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[warm python]

So here I suppose I want to setup an isomorphism b/w $V/ null (\phi)$ and F$?

[soft zealot]

wai
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[warm python]

$v+ null( \phi) \to \phi(v)$. Every non-zero output, is a muliple of some some $\phi (v)$ so we're doen

[soft zealot]

wai

[warm python]

.close

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[severe robin]

how would I apporoach this question

[tired walrus]

do you know roots of unity

[severe robin]

no I don't recall from my notes

[tired walrus]

💀

[vital crag]

I guess use 1 = exp(2pi i n) for any integer n and take 1/6 power

[severe robin]

soz

[tired walrus]

2pi i n

[vital crag]

pain

Thank you

[severe robin]

I'd start with 0 for multiples of 6 right

[vital crag]

I don't know what you mean

Just write out 6 roots

[severe robin]

for any integers n, is it limited to those under 6?

[vital crag]

No. Any integers

[severe robin]

ah okay

.close

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[ebon sigil]

Dudes was the formula like this

[ebon sigil]

Because I don't think 1 can be in root

like wait my language barrier. I don't think there is a root of 1

[final tangle]

wdym by root of 1

oh, 1 sec, you have the m and n swapped around

also the 1th root of a number is just itself

[ebon sigil]

I found it, friend

[final tangle]

$a^{\frac{\blue{m}}{\red{n}}} = \sqrt[\red{n}]{a^{\blue{m}}}$

[soft zealot]

ραμOmeganato5

[ebon sigil]

Yeah I know

I found it thanks :))

You're awesome

.close

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[ebon sigil]

Me again. I'm not sure if that's the right way to find where x belongs.

[ebon sigil]

Hold up hold it be just that, instead of x>4 x- 4 NOT EQUAL 0, which makes x NOT EQUAL 4, but the rest is fine..?

Is that it?

But x can also be a negative number since it's in a root of an odd number...

[versed cargo]

Can i see the question?

[ebon sigil]

Yeah it's this one, in the circle

[versed cargo]

Is that supposed to be an inequality?

[ebon sigil]

My first language isn't English, by inequality do you mean the way 3/x-4?

[versed cargo]

Oh i get it

We need to find the domain?

[ebon sigil]

I need to find the Definition set of X in each equation, and find where X can be on the line of numbers

It's a funny thing, languages

Wait I'll write what I need and send u

[versed cargo]

You are not allowed to divide both sides of an equality by x

[ebon sigil]

Oh I see

What do I do then

[versed cargo]

Factor out x

[ebon sigil]

This one right

[versed cargo]

Yep

[ebon sigil]

How do I factory it

Factor

Factor out

[versed cargo]

x^2-3x=x(x-3)

[ebon sigil]

Ah I see what this means now, I forgot about this

Wait but my answer is to stay the same?

No wait

X is bigger than 3, it isn't 4, but also bigger than 0

[versed cargo]

Ok answer this:

[ebon sigil]

I have a thought that because there is an odd root X can also be a negative number