help-27

[fringe patio]

,rccw

[uneven coral]

AB = BC can be proved by congruency of ABP and CBS

[unique canyon]

Try to prove Congruency

[uneven coral]

Same with AR = RC, congruency of RAB and RCB

[fringe patio]

ohh is it asa

[unique canyon]

Hint : ||ASA||

[uneven coral]

Yes, ASA for (a) and SAS for (b)

[fringe patio]

what about

b to AR

[uneven coral]

b) part?

[fringe patio]

the part after AR=RC

[uneven coral]

Yeah, SAS between RBA and RBC

Oh wait, didn't see that

[fringe patio]

nvm its 4.5

if they are congrunet right

[uneven coral]

No wait

BS is 4.5, how could BR be 4.5

[fringe patio]

i meant B to AR

perpendicualr distance

[uneven coral]

Yes, and the perpendicular on B to AR is BR

[fringe patio]

BR isnt perpendicular tho

I think it means add a new line from point B that will be perpendicular to the surface of AR

and find that oength

[uneven coral]

It is right angled on AC, but yes, even I'm a bit confused

Oh wait

Do you have the answers?

[fringe patio]

i dont

[uneven coral]

I see

We can simply find BR though, IG that's what they meant, the other things would be kindw complicated to find, and this question doesn't seem all that complicated

[fringe patio]

isnt it more like this

[uneven coral]

Could be, but that'd be insane to find

[fringe patio]

if they're congruent ARB and BRC then it would just be 4.5 cuz they are both altitudes

[uneven coral]

Yeah you're right

Oooo missed that

Yeah that makes more sense than BR

[fringe patio]

wait there one more can u help me with

,rccw

[uneven coral]

I'm here

[fringe patio]

any ideas?

cant find anything 😭

[uneven coral]

You've already done a lot lol

[fringe patio]

yeah but idk how to find theta

[uneven coral]

There has to be something, gotta think

Theta is EAC or EAD?

[fringe patio]

eac

[uneven coral]

Okay

That AED is being such a headache

Sorry man, I can't tackle it

[fringe patio]

alr algs

<@&286206848099549185>

[devout snow]

@fringe patio Has your question been resolved?

[north grail]

Try creating an internal angles line or what they use to teach that and bend it.

[devout snow]

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[fringe olive]

Can somebody solve me all of these? I am still trying how it works.

[willow helm]

no

We can help you solve them

[fringe olive]

Ok

Just one

This one

[willow helm]

Not for you

[fringe olive]

Btw this is a past papers

Not an exam

So I basically just try to multiply them based on matrix right

[-1 . 5 2. 1] = [ -5 , 2]

[willow helm]

[fringe olive]

😭

I'm cookrd

Pray for me

[willow helm]

Do you know how to multiply matrices

[fringe olive]

Yes

But not really

So I should say no

[willow helm]

I see

[fringe olive]

Can you give me an example

So I can try working on it

[willow helm]

I will show you how to multiply 2x2 matrices

The process is the same for any size though

So pretty much

You start with the top left element

And multiply it by the top left element of the second matrix

a * e

Then, you move right on the first matrix and down on the second matrix

So b * g

Leading + throughout

And you continue this process until you run out of elements

Then you move to the second row on the first matrix

And repeat the same process

So left most element on the second row, c, times top left element in the second matrix, e.

[fringe olive]

OOO i see

[willow helm]

Then move right -> d and down -> g

[lapis quarry]

Multiply the first row with the first column and add up the elements, first row with the second column, first row with the third column, etc. and when you did everything for the first row you move to the second row and multiply it with the first column, then with the second column, etc.

[willow helm]

You have to do this for every row and column

[fringe olive]

What about this tho, this does not have a 2x2 but a 1x1 horizontally, 2x2 and 1x1 vertically

[lapis quarry]

even easier then

just stop after the first step

[willow helm]

It doesn't matter, as long as the amount of rows in the first matrix equals the amount of columns in the second matrix

Otherwise you would run out of elements to multiply somewhere

[
[-1, 2] \cdot
\begin{bmatrix} 3 & -1 \ 2 & 4 \end{bmatrix} \cdot
\begin{bmatrix} 5 \ 1 \end{bmatrix}
]

[woven radish]

USS-Enterprise

[willow helm]

We start with the first two

[
[-1, 2] \cdot
\begin{bmatrix} 3 & -1 \ 2 & 4 \end{bmatrix}
]

[woven radish]

USS-Enterprise

[willow helm]

As we said, left most element in the first row of the first matrix TIMES the upper most element in the first column of the second matrix

(-1) * 3

PLUS

The next element in the first row TIMES the one lower element in the first column

2 * 2

And that is our first element of the resulting matrix

(-1) * 3 + 2 * 2 = 1

Get it?

[fringe olive]

Yes

[willow helm]

Okay

Then we move to the second column on the second matrix

And we begin the same way

[fringe olive]

So it's [-1 . 3 + 2.2 = [1]

[willow helm]

(-1) * (-1) + 2* 4

[fringe olive]

Ohh

So it must be 9

[willow helm]

Yes

[
[-1, 2] \cdot
\begin{bmatrix} 3 & -1 \ 2 & 4 \end{bmatrix} = [1, 9]
]

Right?

[woven radish]

USS-Enterprise

[fringe olive]

Yes

[willow helm]

If you move columns like here, the next element goes to the right

If you move rows (we didn't because the first matrix only has 1 row) you move down on the resulting matrix

So in the end the result matrix is the size m * n, where m is the number of rows in the first matrix and n the number of columns in the second matrix

[fringe olive]

Wait what about the (5,1)?

[willow helm]

We are getting there

Now we have

[
[1, 9] \cdot \begin{bmatrix} 5 \ 1 \end{bmatrix}
]

[woven radish]

USS-Enterprise

[willow helm]

How do we do this then

[fringe olive]

So it's [14] ?

[willow helm]

Exactly

[fringe olive]

That's the final answer?

[willow helm]

Yes

[fringe olive]

TYSM!!

[willow helm]

No problem

[fringe olive]

LOVE YOU BROTHER ( NO HOMO )

[willow helm]

🙂

[fringe olive]

.close

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[low parrot]

A and B play the following game: A writes down
either number 1 or number 2, and B must guess
which one. If the number that A has written down
is i and B has guessed correctly, B receives i units
from A. If B makes a wrong guess, B pays 3
4 unit to
A. If B randomizes his decision by guessing 1 with
probability p and 2 with probability 1− p, deter-
mine his expected gain if (a) A has written down
number 1 and (b) A has written down number 2.
What value of p maximizes the minimum pos-
sible value of B’s expected gain, and what is
this maximin value?

[low parrot]

i thought of first saying P(1) = P(1|B guess)P(B guess) + P(1|B not guess)P(B not guess)

is it a right move?

[pseudo basin]

are you doing a) or b)

also just to clarify... how much is B supposed to pay A on a wrong guess?

[manic sedge]

B pays 34 units to what?

3 , or 4 or 34

[low parrot]

what?

ohhh sorry

3/4 it is

[low parrot]

a

[pseudo basin]

ok so B gains €1 if he guesses correctly and the number was 1, €2 if he guesses correctly and the number was 2, and on any wrong guess he loses €0.75

[low parrot]

yes

[pseudo basin]

so a) is the scenario in which A's number is 1. which means that B's payouts are:

• +€1 with probability p
• -€0.75 with probability (1-p)

find B's expected payout in this case

no probability or anything but expectation aka expected value

[low parrot]

lemme use event G_B as gain of B

[pseudo basin]

G_B is not an event

it will be a random variable

[low parrot]

i mean r.v

sry

so like

E[G_B | A = 1 ] = p + q*-.75 right?

@pseudo basin

[pseudo basin]

p×1 + (1-p)×(-0.75) more like

[low parrot]

ye lol same thing

q = (1-p) btw

oh lemme just say 1-p to simplify

it'd be like (7/4)p - 3/4

E[G_B | A=2] = (p)(-.75) + (1-p)(2)

2 - 11/4p it'd be

[pseudo basin]

yes now you need to find the value of p which equalizes these

[low parrot]

i shall find the maximin huh

p* = 11/18 ig

yup

alright thx Ann u cool

.close

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[dry harbor]

hey can someone help me understand optimal linear prediction?

[dry harbor]

i know you have to draw a line in the graph, but no idea how you know how to draw said line

[thorny bloom]

can u give an example problem

[dry harbor]

the book doesn't have like excersises but i can show an example that i'm trying to figure out how they did that?

[thorny bloom]

sure

[dry harbor]

do you mind that it's upside down?

[thorny bloom]

is it machine learning

[dry harbor]

i don't think so? no idea what you mean by that tbh

it's like you got data for all these points and then you need to predict what a new point will have when it only has the X known, based on what all the other points are

it could be, but machine learning is not mentioned anywhere in the book

oh zlr thanks

.close

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[pure stone]

Prove $\int_{0}^{\pi} \frac {1}{\sqrt[3]{\sin x}} ,dx$ is convergent

[woven radish]

Carbonara

[pure stone]

$\int_{0}^{\pi} \frac {1}{\sqrt[3]{\sin x}} dx$ = 2 $\int_{0}^{\frac {\pi}{2}} \frac {1}{\sqrt[3]{\sin x}} ,dx$ < 2$\int_{0}^{\frac {\pi}{2}} \frac {1}{\sqrt[3]{x-\frac{x^3}{3}}} ,dx$

[woven radish]

Carbonara

[pure stone]

what should i do next

[sand quarry]

i feel like you are overcomplicating it a bit

hint: utilise the fact that sinx \approx x for small x and use the limit comparison test

[pseudo basin]

if you want something exact ig you can use sin(x)>=2x/pi for x in [0,pi/2]

[pure stone]

oh

that's it

thanks

.close

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[feral agate]

how do you use LCT for integrals

[sand quarry]

the same way you use them for infinite sums

[devout snow]

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[floral lodge]

.close

[devout snow]

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as this action is irreversible, and this channel may abruptly lock.

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[floral lodge]

.reopen

I am having issues understanding question 15 and 16

[soft umbra]

Gotta claim another one

Since you’ve deleted the original message

[floral lodge]

aww 🙁

[soft umbra]

^^

[floral lodge]

I deleted as it was badly formatted

[soft umbra]

It’s alright, the next one will be better

[devout snow]

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[verbal comet]

help 🥺

[violet wind]

Start by focusing on the limit

What does that tell you about the polynomial?

[verbal comet]

dont remember 🥺

[fossil locust]

what's the limit of $\lim_{x \to -\infty} \frac{5x^2 + 2x + 1}{x^2 + x + 2}$ for example?

[woven radish]

south

[verbal comet]

lim 5x^2 / x^2 ?

[fossil locust]

yep!

[verbal comet]

so lim f(x) / x^2 = 3?

[fossil locust]

yes

[verbal comet]

I see.

[fossil locust]

importantly, if $\lim \frac{f(x)}{g(x)}$ is not $0, \infty, -\infty$

[woven radish]

south

[fossil locust]

that means $f(x)$ and $g(x)$ must be of the same degree

[woven radish]

south

[devout snow]

@verbal comet Has your question been resolved?

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[nimble turret]

Yo I got a doubt

[steel tinsel]

!da2a

[devout snow]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question!

[sharp garnet]

just ask

[nimble turret]

Ye it was loading

[steel tinsel]

,rotate

[nimble turret]

So the thing is

The pt 2a,0

[nimble turret]

Will have 3 coincident normals

Right?

Lemme explain this

Parabola normal equation

Got relationship btw m values so that the cubic has 3 roots

Hence 3 normals pass from a pt

Got this graph

That shows number of normals that can pass through a pt in each region

Caught up?

Well the question is what is the equation of 3 coincident normals that pass through pt 2a,0

Like tell me how to get it

Also the 3 coincident normals are supposed to pass through vertex of parabola and pt 2a,0

But not any other pt on axis

Like only 1 of those 3 will pass through rest of em

How's that happening

Please help needed

[supple knot]

!original

[devout snow]

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.

[nimble turret]

Ye no there was no problem given

Just explanation

And I had doubt in that

And I sent the related theory that was given

Like a parabola assumed y2=4ax I think

At the pt 2a,0

Will have 3 normals passing through it

Which will be coincident

As a rule

[supple knot]

explanation of what problem

[nimble turret]

Only theory was taught about no. Of normals passing through diff pts on parabola

Like I think the only pre required info was the parabola is y2=4ax I think

I don't think any other assumptions were taken

Do you want me to give a rundown on what is going on in the image?

[rare kernel]

U can't get equation of 3 normals from 1 pt

[nimble turret]

Well

We got it other way around I think

We first got the equation of normal

Then gave condition for 3 solution

[rare kernel]

U can get the slopes of the normals

[nimble turret]

Yea but they are coincident

The equation 27ak2 ....

[rare kernel]

Could u explain ur question again? Ur typing in so many lines I xcant understand

[nimble turret]

From the pt 2a,0 3 coincident normals of parabola y2=4ax pass through,
Somehow the 3 coindent normals do not satisfy any point on the axis other then 2a,0

Like it's there but it aint

[rare kernel]

And why do u think they will satisfy any other point on the axis?

[nimble turret]

Cuz well they pass through the axis

One of the normal is the axis

And if they are coindent

Then others must be same no?

[rare kernel]

Huh coincident

[nimble turret]

Uh yea

[rare kernel]

Why will the parabola have 3 coincident normals..

[nimble turret]

Yea I'm confused about that part too

But as far as ik

[rare kernel]

Is there an original question

[nimble turret]

The equation would be M3=0

Ye no there aint

[rare kernel]

How is that an equation

[nimble turret]

The image I sent is all

Ye well I think there was something there but it got simplified to this

I know I am not being helpful at all

[rare kernel]

All I can say is that the h>2a condition is correct for 3 normals of a parabola to exist

[nimble turret]

Well lemme tell what I understood from class

[nimble turret]

After doing that

M1 m2 when put in orignal function and multiplied

Will give negative result

So we did that and put inequality less than. 0

So we got more accurate discription of when 3 normals

[rare kernel]

Yes

[nimble turret]

Ye so from that we did h=2a

And mess started

[nimble turret]

Like when that condition follows we get 3 distinct normals

[rare kernel]

Why h=2a?

[nimble turret]

No like if we take h=2a

There are supposed to be 3 normals

Just not distinct

Like just tell me this

[rare kernel]

Where did tangents come from

[nimble turret]

Yo mb

[rare kernel]

We are using equation of normal right

[nimble turret]

I have been tripping btw normals and tangents whole day

I meant normals

[rare kernel]

The condition we are solving for is all for normals

Ohk

[nimble turret]

Tell me this

Like give me example of a coincident line

Like x2=0 that's a line right

Coincident?

[rare kernel]

U can say that I guess but if they are coincident then how can u classify them as different normals?

[nimble turret]

Ye that's why the condition won't hold

Like it would be 0<0

[rare kernel]

Yes obviously it won't.. the condition is for 3 distinct normals we are not getting 3 distinct normals

[nimble turret]

I don't think I can recall exactly what the teacher said

So I'll just ask in general way

X2=0

[rare kernel]

Btw are u preparing for an exam

[nimble turret]

Is equation of lines right

[rare kernel]

What is x2

[nimble turret]

Variable like x of x axis

X sqaure

[nimble turret]

Like it's far away

Yo

I think I have a way to explain

But I gotta restart it's all too cluttered I don't wanna waste another age explaining to someone again

.close

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[floral lodge]

Why is the answer to this no? The answer sheet gives one contradiction but the question asks if there is some linear transformation given $x_1,x_2\in V$and $y_1,y_2 \in W$ such that $T(x_1) = y_1 $ and $T(x_2) = y_2$

[woven radish]

Pen

[floral lodge]

If V is a subspace of W, then surerly I can map the vectors to themselves?

or if V and W are of the same cardinality (is that the right term here?), mapping each x_i to y_i

[drifting sierra]

Why are you making these assumptions?

[floral lodge]

Because there is no restriction on V and W

and it asks if there is some linear transformation and not for all transformations

[drifting sierra]

Yeah but you're not the one choosing

[floral lodge]

Then how else am I supposed to answer it?

If it said for all linear transformations given V and W, then sure

but that does not seem to be the case

[drifting sierra]

Maybe there's more context you're not showing here, but this reads as "for all x_1, x_2 in V and [...], there exists [...]"

[drifting sierra]

I mean, just take V = W = R

x_1 = 1, x_2 = 2, y_1 = 1, y_2 = 3

T(0) must be 0, and T(1) = 1, so T(x) = x

But then T(2) =/= 3

[floral lodge]

Okay, that is a counterexample to those values

if V = W = R I can let x_1 = 1, y_1 = 1,x_2 = 2, y_2 = 2

[drifting sierra]

It's a counterexample to the statement

[floral lodge]

My bad, yes

Hmm

Why is it a counterexample to the whole statement and not just those values?

[drifting sierra]

What you don't get is that you cannot just pick the values you want for it to work

[floral lodge]

Why is $x_i = y_i$ not a valid choice here?

[woven radish]

Pen

[drifting sierra]

The statement is "for all [...], then there exists [...]"

It is a valid choice, but not the only choice

[floral lodge]

right, it is not the only choice but it is a choice

wait

Nel, do you mean that for ALL values in the set

[drifting sierra]

If all valid choices made it work, then the statement would be true

My counterexample is a valid choice, yet it doesn't work

[floral lodge]

so any random $x_1, x_2$ in V has to satisfy the conditions given any random $y_1,.y_2$ in W

[woven radish]

Pen

[floral lodge]

ahh okay

I see my mistakes

Nel, thank you for being patient enough with me

[drifting sierra]

If the statement said "affine transformation" instead of linear, then it would be true

Or if it only had one value x and one value y

(the transformation being linear means there's already an implicit pair (x,y) that must work, namely (0,0))

[floral lodge]

Yep

I need to take a step back and re-tackle the phrasing of some of the questions, I am not reading it correctly

Thank you so much, Nel!

.close

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[viral lynx]

In Exercises 121–124, determine whether the sequence is monotonic and whether it is bounded.

[heavy current]

well, what do you think?

[viral lynx]

This is what the solution says, but I dont see why

[polar chasm]

why what part?

[heavy current]

which part of the solution is the confusing part?

[viral lynx]

In order for it to converge, it needs to be increasing and have an upper bound, right?

[heavy current]

sure

[viral lynx]

But its not increasing right?

[heavy current]

why not?

[viral lynx]

Cause its negative

[heavy current]

write out a few terms of the sequence

[viral lynx]

So its getting smaller as n increases

[heavy current]

try writing out some elements :p

I think you'll change your mind

[viral lynx]

1 - 1/2 = 1/2

1 - 1/3 = 2/3

Oh

😅

Ok I see your point

Thank you!

❤️

.close

[heavy current]

[devout snow]

Channel closed

Closed by @viral lynx

Use .reopen if this was a mistake.

[viral lynx]

Also wow help from higher, I am honoured

[heavy current]

https://tenor.com/view/shy-blush-sawamura-spencer-eriri-anime-saekano-gif-21117029

[devout snow]

Available help channel!

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[azure sleet]

How do I make a help channel bruh

[azure sleet]

Guys

[silk terrace]

You have successfully made one

Ask away

[azure sleet]

Can someone help me with discrete mathematics sets and relations

[faint hearth]

send any mathematical questions you have here

[cold haven]

No need to ask — there’s always going to be someone that can help.

[faint hearth]

which question do you have a problem with?

[drowsy minnow]

all

[silk terrace]

!status

[devout snow]

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

[dusky crown]

How do I factorize a polynomial of 4?

[drowsy minnow]

!1

[silk terrace]

Do you know how one proves a biimplication?

Biimplication= if and only if statement

[drowsy minnow]

yes

[silk terrace]

Okay, can u tell me

[drowsy minnow]

if the two statments are true its true

1 <--> 1 = 1

and if false too

[silk terrace]

A biimplication is really the and of two implications
So you have to prove two statements

p => q and q => p

So for 6-8

You have to prove that

1. If power set A is a subset of power set B then A is subset of B
2. If A is subset of B then power set A is subset powerset B

[drowsy minnow]

yes ik

how do I prove it

[silk terrace]

How will you go about proving 1.

[drowsy minnow]

idk

[silk terrace]

If you take any element in A

And make a singleton set with it

It will be in P(A)

Makes sense?

And this P(A) is a subset of P(B)

Which means that singleton set is also in P(B)

Which means that element must be present in B

So any element in A is present in B

=> A is subset of B

Tell me if this makes sense

[drowsy minnow]

I get u

but where are the sets

[silk terrace]

What do you mean where

[drowsy minnow]

like how will I prove that if I dont have the sets

like set a and b

@silk terrace

[devout snow]

@azure sleet Has your question been resolved?

[silk terrace]

I mean assuming that A and B are not empty sets

They will contain some element x

Rest of the argument follows

For any arbitrary set

[devout snow]

@azure sleet Has your question been resolved?

[devout snow]

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[potent pike]

Im confusingmyself here

I don't think i really understand what im looking for?

i think i dont really understand the question

depth is referring to height correct?

[woven vale]

yes, your diameter looks correct

[potent pike]

ahhhh

[woven vale]

this seems pretty similar to the cylindrical one you did. how about tracing what you did in that problem, so you can figure out what is confusing you here?

[potent pike]

I think im confused with depth

i know how to do it in theory but like the terms are confusing me a little

let me try it once more

@woven vale radius isn't a constant for this right?

so i do need dr/dt

[woven vale]

this is fair, and i honestly think its kinda mean that they used depth and height in the same problem

think of depth as the height of the cup, or how "tall" the liquid will look inside the cup

if this helps

[woven vale]

radius is not constant, correct. if you imagine looking from above at this cone, you will see a circle of water that gets smaller and smaller, right?

thus the radius of this circle is changing

i like to start by identifying whats going to change and what's not, within the formula for volume or w/e

then taking derivatives is a little easier

[potent pike]

yes thats what i was assuming aswell

[potent pike]

im so confused still with what depth is

can i ijust assume that its height

[woven vale]

yes

[potent pike]

okay let me try to solve it now then

[woven vale]

also just wanted to point out that this isn't correct :(

[potent pike]

oh no

h = (v/pi(r^2)3)

idk i dont think i can solve thid question rn

im very confused

[woven vale]

okay, don't worry about isolating variables right now

[potent pike]

no i did that

but

[woven vale]

focus on taking the derivative properly, then you can rearrange

[potent pike]

the erst of the qeustion

i don't understand

[woven vale]

what is the "rest"?

[potent pike]

okay let me try and solve it again

[woven vale]

okay sounds good

if you're not sure what part of the word problem means, or how to apply it, you can circle it and send that too

[potent pike]

is dh/dv right so far

@woven vale

im loset

this is one hard question

so lonely

😭

[devout snow]

@potent pike Has your question been resolved?

[devout snow]

Channel closed

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[viral lynx]

I got the first 8 terms bit but how do i calculate the sum

Isnt it infinite?

[faint gorge]

have u heard of geometric series

[viral lynx]

Yes we learned them today but i dont fully understand them

[faint gorge]

do you know what a geometric sequence is

[viral lynx]

Same thing

[faint gorge]

no

[viral lynx]

I mean same like we learned it today but i dont fully understand it

[faint gorge]

it's when you get the next term by multiplying that with the previous one by a common ratioo

[viral lynx]

Ah ok

[faint gorge]

1, 2, 4, 8, 16...
all terms are multiplied with the common term 2

that is a geometric sequence

and now if you were to add up all these numbers, you get now a geometric series

[viral lynx]

But adding them would get infinity no?

[faint gorge]

not necessarily

[viral lynx]

Cause the sequence is infinite

[faint gorge]

imagine you have 1 chocolate bar

[viral lynx]

Mmmm

Yum

[faint gorge]

you can divide it into half, then 4 parts, then 8 parts

etc.

doesnt mean you get infinite chocolate now, does it

[viral lynx]

Ok I see

I dont really know how to answer the question then

[faint gorge]

have you learnt a formula for geometric series

[elfin meadow]

I have a geometry test in 3 days 🙏🏻

[viral lynx]

Congrats

[supple knot]

!occupied

[devout snow]

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).

[viral lynx]

Nope 1 sec ill google it

Oh yeah I remember this from the lecture

[faint gorge]

this is btw how you may imagine the chocolate bar

[elfin meadow]

Oh

[faint gorge]

yea it was for finite sum, surely

[viral lynx]

What is the ratio and how do I get it

[faint gorge]

write down soem terms

[viral lynx]

-1/4

1/16

-1/32

1/64

[faint gorge]

where is the negative come from and it starts at n=2

[viral lynx]

Oh oops I was doing the wrong question

The one before was -1^n

1/4
1/16
1/32
1/64

[faint gorge]

if you look at the terms what did you do

what does (1/4)^n suggest doing

as n increments

[viral lynx]

In terms of if its increasing or decreasing?

[faint gorge]

no, arithmetically

what are you doing in every new step

how does 1/4 differ from 1/16

[viral lynx]

The denominator is increasing by a factor of 3 then 4 then 5

[faint gorge]

no

then you would have 1/4, 1/12, 1/48, etc ...

[viral lynx]

Hmmm

Ah

[faint gorge]

you are saying that: 1/4, 1/(4x3), 1/(4x3x4), ...

[viral lynx]

The denominator is 2^(n+1) every time

[faint gorge]

it's literally here what it is

[viral lynx]

Like each time it adds 1 to n

[faint gorge]

1/4^n

1/4 then 1/(4x4) then 1/(4x4x4), etc.

[viral lynx]

Ah yeah

So the ratio is 4^n?

[faint gorge]

1/4

[viral lynx]

Hmm

But its not r^2 its 1/r^2

[faint gorge]

r is just a variable

r could be 2 or 1/4 like in your example

[viral lynx]

I still dont understand why its 1/4 sry

Like if the equation was (1^n)/4 would the ratio still be 1/4

Or (1/4)^n

[faint gorge]

you are multiplying by 1/4 everytime not (1/4)^n

as n increments, you always multiply by an additional 1/4

[viral lynx]

Ahhhh I see

Okokok I think I know how to do it now

You explained it really well!

[elfin meadow]

Fr i was watching all the time

[faint gorge]

so since z=1/4 we have 1/4 < 1 so it converges

[viral lynx]

Ahhh ok

I dont think we learned this today

Or maybe we did lol idk

But im gonna try some examples

[faint gorge]

you can derive it by taking the limit of the formula

[viral lynx]

I will just learn it instead 😅

But thanks again!

❤️

.close

[devout snow]

Channel closed

Closed by @viral lynx

Use .reopen if this was a mistake.

[elfin meadow]

Guys

[faint gorge]

learning by proving

[elfin meadow]

Is the 80-20 rule work

[elfin meadow]

I had a question

Wait

[faint gorge]

you are free to open a help channel

[elfin meadow]

Ok

[viral lynx]

.reopen

[devout snow]

✅ Original question: #help-27 message

[viral lynx]

This is what i got when trying to sub in the formula

[faint gorge]

show the formula and then compare it to your sum and start index

[viral lynx]

a_1 is 1/4

r is also 1/4

n is 2

[faint gorge]

your first term is not 1/4

what is your first term, when n=2

[viral lynx]

Ohhhhh

Hmmmm

[faint gorge]

why did you square

it's an infnite sum

[viral lynx]

Oh

[faint gorge]

n goes to inf

[viral lynx]

So n isnt the n = 2 under the sigma?

Thats confusing

But I see what you mean

So (1/4)^infinity tends to 0

[faint gorge]

yes

[viral lynx]

Ok I see

Finally got the answer they wanted

Thank you!! Again

❤️

.close

[devout snow]

Channel closed

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[devout snow]

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[zenith owl]

how can i demonstrate the commutativity of the multiplication? a times b being equal to b times a

[cold haven]

you could always use rectangles (width 𝑎 and height 𝑏) and rotate the rectangle by 90º (width 𝑏 and height 𝑎). The area of the rectangle wouldn’t change just because you rotated it.

If you want a rigorous proof, you’d probably need peano axioms and induction.

[zenith owl]

i thought about that but i can't think of a way to demonstrate that the area of any rectangle is width times height, like why the dimension meter multplicates with another one and we assume meter squared represents an area

ik these are the basics but i can't find demonstrations for this type of stuff lol

[devout snow]

@zenith owl Has your question been resolved?

[celest frigate]

Consider two numbers $a,b$ WLOG $a\geq b$

We assume they are not equal
$$a + a + a + ...\text{b times} \neq b + b + b+ ... \text{a times}$$

Since $a>b$, we write ($a=b$ case is trivial)
$$(a-b) + ... \text{b times} \neq b+b+... \text{a-b times} $$

We repeat the procedure finitely many times till the two numbers are equal, and then it becomes trivial $c\times c=c\times c$

[woven radish]

Adhi

[celest frigate]

@zenith owl is this satisfactory?

This is the best I could think of using the least vague axioms

we only used equals subtracted from equals are equals and also we have to repeat the procedure for whichever is lesser than the other

[zenith owl]

i didn't understand it quite well

i got lost on the second step

[devout snow]

@zenith owl Has your question been resolved?

[devout snow]

@zenith owl Has your question been resolved?

[gloomy aurora]

if you consider commutativity for natural numbers as @celest frigate showed above, then you can show it for rational numbers if you consider them as ordered pairs $(p,q) \in \mathbb{Z}\cdot\mathbb{Z^*}$

[woven radish]

Annie Maqionde

[gloomy aurora]

And define equivalence relations etc on them

for real numbers, dedekind cuts might be useful

[devout snow]

@zenith owl Has your question been resolved?

[devout snow]

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[spiral plover]

Q 10 im not sure if the way i am answering the question answers it, if u know what i mean

[spiral plover]

Drew the graphs

Actually i think i did it graphically

[hidden star]

It says algebraically, right? Also, it seems cut.

[spiral plover]

O sorry

Mb

Is it something to do with gradients?

[hidden star]

There's no need to involve gradients, I believe.

You can just make use of the equation of the line, substituting it to the eq of the circle

Aren't you able to get the value of y in terms of x? Use it

[pseudo basin]

think about how it may manifest algebraically that the line and circle would be tangent (ie have exactly one point in common)

[spiral plover]

So i should find the contact point first?

[hidden star]

You have to prove it first that it's tangent before finding the contact point.

[spiral plover]

O well i found the contact point

Something ti do with right angles?

Perpendicular

[pseudo basin]

ok i feel like there has been a "missed forest for the trees" moment

[pseudo basin]

can you show your work

[spiral plover]

Ill quickly rewrite it

Is it ok if i hold this channel while i eat shouldnt be too long

[pseudo basin]

ok so you reduced the thing to a quadratic in x

and you solved it, and -- importantly -- you found that its discriminant is zero and that it has only one solution.

[devout snow]

@spiral plover Has your question been resolved?

[spiral plover]

Sorry im back

[spiral plover]

Is this enough evidence to prove that its a tangent

[hidden star]

Yes

[spiral plover]

Ok thanks

.close

[devout snow]

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[devout snow]

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[lost laurel]

I don't. get this

[tall knoll]

what don't you get?

[devout snow]

@lost laurel Has your question been resolved?

Channel closed

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[tall knoll]

https://tenor.com/view/frodo-keep-your-secrets-alright-then-lotr-gif-12907121

[devout snow]

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[glossy dew]

pretty sure its B..

[wraith horizon]

why do you think it's B

also which opne is B

[glossy dew]

top right

i didnt see the option markings were removed sry

$\binom{55}{r-1}2^{55-(r-1)}(\frac{1}{3})^{r-1} = \binom{55}{r}2^{55-r}(\frac{1}{3})^{r}$

[glossy dew]

using this as the equation to solve for r

answer would be of the form r, r+1

[runic prawn]

use calculator to check ur answer

and the other options

coeff of $x^r$ is $\binom{55}{r}2^{55-r}(\frac{1}{3})^{r}$

[glossy dew]

yeah but i don't want to just trip over on the same mistake in the calculator

[runic prawn]

just plug in values of r

let's say on desmos

[glossy dew]

its just the off-by-one error thats ticking me off

[runic prawn]

u can see the formula so it's hard to make a mistake

do u need like a package to use \binom

[glossy dew]

dont think so

i just wanted extra validation from someone

so

[woven radish]

sheaf of stacks

[runic prawn]

tbh a calculator might have difficulty handling this

[devout snow]

@glossy dew Has your question been resolved?

[full trellis]

I agree

[glossy dew]

well thanks

.close

[devout snow]

Channel closed

Closed by @glossy dew

Use .reopen if this was a mistake.

[full trellis]

@glossy dew Maple also agrees

[glossy dew]

man what IS that ✝️

eldritch calculator

[full trellis]

You can see the two equal coefficients in the expansion!

That's all that matters

@glossy dew It is 7th and 8th btw, not 8th and 9th

Idk what B is in your picture

[glossy dew]

no its 8th and 9th in what you showed too

the computer decided to troll me by placing the x term at the start

[full trellis]

Oh yeah I forgot about x^0 being the first term

Whoops

[devout snow]

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[modest shadow]

Hello. Suppose we wanna solve the equation: z^6 = -1, and graph the solutions in the complex plane. now, i dont rlly remember the solutions, but my question is: if we find the argument and (i think) the root of unity, can we just write it in the form of re^(i(Arg(z)+2kπ) for k=0,1,2,...,5 and graph it easily like that? I ask this because our professor hasnt taught us the exponential form and we have to evaluate the trig functions and get it down to the form of a+bi. Or is the exponential form missing smth?

[river rune]

I'm quite sure that the solutions are ||evenly spaced out around a circle of unit length||, so I'm not sure how correct that expression is on first inspection.
the idea of using roots of unity might apply but in a slightly different way, I believe?
(do wait for other helpers though. I am not confident in this.)

[vital edge]

the good thing about equations of the form z^n = k is that you only need to find one solution

[vital edge]

That is correct

Now observe that if points are evenly spaced around a circle, then you can get from one to the next by multiplying a root of unity

[modest shadow]

we need only find the argument and the absolute value of one of the roots right?

[vital edge]

Yes

[modest shadow]

ohhh i didnt know that

[vital edge]

Just write the RHS in the e^iθ form

Well r is 1 here

[modest shadow]

and theta is pi

[vital edge]

I can never remember but I think so yea

[modest shadow]

hmm so we only need to find how many degrees the root of unity rotates smth upon multiplying then right?

[vital edge]

That's easy no, the principal nth root of unity rotates it by 360/n degrees

Cuz applying it n times takes you to 360

[modest shadow]

true

[vital edge]

Also Lute feel free to talk if you have any comments

You don't need to step back lol

[modest shadow]

i think i understand now

thank you both

.solved

[devout snow]

Channel closed

Closed by @modest shadow

Use .reopen if this was a mistake.

[sand quarry]

Oh its fijo

[modest shadow]

im having trouble in class cuz i missed a week lol

[sand quarry]

Btw just as a general tip

For complex numbers

If it involves multiplication/division you are better off using the polar (exponential) form usually

[modest shadow]

our prof doesnt allow us to use it in exams lmao

[sand quarry]

If it involves addition/subtraction you are better off using the rectangular form

[modest shadow]

ohhhh okeoke

thats acc useful to know

thanks

[sand quarry]

I wonder what caused that hmm...

[devout snow]

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[rotund token]

why isn't the answer of this the integral of x^2-3 added with the integral of 1

between the bounds of -2 and 2

[thick ledge]

What do you mean by with

Addition? Subtraction?

[rotund token]

addition

[thick ledge]

Why addition?

Also lots of things are wrong with your choice (including this)

[rotund token]

unless im interpreting this wrong

it seems intuitive to just add them both

[thick ledge]

I presume the question asks for area of shaded region?

[rotund token]

between the bounds

[thick ledge]

How?

[rotund token]

yes

[thick ledge]

Youre adding the strip twice

[rotund token]

I thought that the integral was just the area of the function relative to the restriction of the x-axis

like how I shaded in the two graphs above

[thick ledge]

Integral is signed area.

When u take - 2 to 2 bound for g(x) = 1 you are including the unshaded patch which ive outlined with blue

Moreover, integrating -2 to 2 for the parabola will most likely give u a negative answer

Since again, integral is signed area, not just area

You will have to do quite a bit of work if youre integrating wrt x

[rotund token]

so the answer is just the integral of g(x)-f(x) between the bounds of [-2,2]?

[thick ledge]

Take some time to process what ive said

Because you're still wrong

Not tryna demotivate you, just want you to think a bit

[rotund token]

I just used the method I just said and it was the correct answer according to the textbook

[thick ledge]

Yes youre right

Im stupid lol

Thats the correct method

[rotund token]

in the example where there isn't the small unshaded patch

am I able to use the method I used before of just adding

because the two functions would just compliment eachother

and not interfere

or no

.close

[devout snow]

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[devout snow]

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[restive river]

How are you meant to integrate 1/(k - root x) dx

[restive river]

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[coarse lodge]

Can anyone explain me the Recursive Relations in Sequence and Series?

[coarse lodge]

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[heavy jay]

is there a faster way to get to the answer or atleast deduce from the multiple choice?

I first did: 2x+2=(3−sqrt(x))^2

then got to the 7−x=6sqrt(x)
​
then (7−x)^2=36x
x^2−50x+49=0

then used the long ass quadratic formula to find the x
x = 49
x = 1

this process is not ideal for me, cause i have an exam with 50 questions for 30 minutes...

[polar chasm]

plug everything in

that's the fastest way

[heavy jay]

ahh

[deep abyss]

x cant be negative

and 49 would give a negative rhs

[heavy jay]

ohhh

thus c

[polar chasm]

yep

[deep abyss]

yh

[heavy jay]

damn

thank you

[polar chasm]

when you square an equation you have to check all the roots in the end, by squaring you can sometimes add some extra solutions, which arent sols

in this case, this extra-root was 49

so this is another reason why just plugging everything in is much faster

[heavy jay]

I see

u guys r very helpful

thank you so much

waaaaaa

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[pulsar relic]

are there any tricks to understanding how to do part b?

i know its asking to find the integral of the marginal-cost function between 0 and 25.

the problem im having is with the u-substitution. (its complicated for me) usually i'd just u-sub everything inside the sqrt. here it doesnt seem to work (?)

looking for general advice on what to do in cases like this and what to look for

[hidden star]

Well, I'm not sure this could help but you could look at the power of q.

[supple knot]

total cost = sum of all marginal cost

But since you have a derivative, it's the integral instead of a sum

[pulsar relic]

when i checked ai it suggested usubbing for just q^3/4. is that what you meant too?

[pulsar relic]

this also means i could solve it by...solving for the sum of all marginal costs by substituting in q=0....25? + fixed costs at the end

[supple knot]

yes. try it and see

[hidden star]

You can try 🥀

[pulsar relic]

dont have time for it rn, but maybe if im stuck on my exam i will use that trick. it does sound right

maybe a follow up to all this would be how to solve, using words if you (anyone) doesnt want to work out the math fully for me to see 😬

[hidden star]

Still part b, I assume?

q^3/2 works like a charm actually, it cancels things nicely.

So, I suggest observing all the possible values you can substitute for u to make the problem look easier. In this case it was: q^3/2

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@pulsar relic Has your question been resolved?

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[split dirge]

Can anyone solve 15 d

I tried this

Nevermind, I got my solution.

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[hushed drift]

hey, i dont get how to solve this page

[supple knot]

oh do you need help on your test?

[hushed drift]

yeah its a practice test but the test is near the same level as the actual exam im about to do

i get circle theorem and that but im not sure what rule i have to apply here

[burnt drift]

work out DBA from the internal angles of a quadrilateral

from there you can get DBC = DCB

[hushed drift]

is that 360 - 90 - 112 - 52?

[burnt drift]

correct

[hushed drift]

i got 106 for the last angle in the quadrilateral, not sure what to do now

180-106?

[burnt drift]

well, if the angle created by DBA + DBC = 180, and DBA = 106, we can see what DBC is

[burnt drift]

yes

[hushed drift]

okay

[hushed drift]

i got 74 for this, i know that two angles in an isocles triangle = eachother, not sure what the next step is, chatgpt is saying i should divide 74/2

[burnt drift]

bad bot
think about it, which two angles are identical?

within the triangle DBC

[hushed drift]

DBC = DCB?

[burnt drift]

and to make sure you understand, why these two?

[hushed drift]

so would it be 180 - 74 - 74?

[hushed drift]

they are from the same chord and have the lines to indicate

[burnt drift]

it's been too long 😂 I forgot what a chord was, but they're the two identical angles in the isoceles triangle because they are not the angle that's sandwiched between the two identical lines

[burnt drift]

yes, x would be that

[hushed drift]

okay i understand thank u!

[hushed drift]

ty

[wild spindle]

Are you done with the problem? If so, you may want to close the channel. If not, I could help because I just learned this so it's fresh in my memory

[hushed drift]

i have other problems i need help with if youre free?

[wild spindle]

Mhm! Ok. I know geometry tho, not functions, although I'm sure I could figure it out

[hushed drift]

here

[wild spindle]

Oh fun!!!

Have you learned secant secant angles?

[hushed drift]

no dont think so

[wild spindle]

Aw dang it that would be the simplest way to solve it

[hushed drift]

i bet itd be easy to learn then?

[wild spindle]

Ah kinda but your teacher may not want you to use it

[hushed drift]

yeah true

[wild spindle]

Basically, a secant is a line segment that passes through the circle such that one end is on the other circle and the other end passes through the circle out the other end. Two secants meeting outside the circle form a secant secant angle. The major arc minus the minor arc divided by 2 equals the secant secant angle

You don't need to know that for your class tho lol

[hushed drift]

lol i kinda get it tho

[wild spindle]

Its very handy. The proof is relatively simple too. Id recommend checking it out if you're interested

Imma try and figure out a different way to solve your problem tho lol

Well, you know angle DOB

[covert delta]

I'm not sure if I'm supposed to intervene since I'm new here so sorry about that but it does say that co = ce

And that makes an isosceles triangle

[wild spindle]

Oh you can intervene it's fine

[covert delta]

So that means finding this angle in red is x

[covert delta]

Oh alright

[hushed drift]

oh i didnt notice yeah ur right

[wild spindle]

And finding angle DCO is 2x

That's why secant secant angles would be so helpful tho. (69-x)/2=x

Wait I'm dumb 😭

DCO is also isosceles because of the radii

So angle DOC is 180-4x

Its simple from there

Hey Xen have you solved it yet? If not, I can give you hints

[wild spindle]

The main thing to remember is that triangle DCO is also isosceles

[hushed drift]

im lost

can u solve it please?

[hushed drift]

i understand this but im not sure what to do with it

[wild spindle]

Oh ok

Well, a triangle adds up to 180

So does a line

So angle DCO =angle CEO plus angle COE

[wild spindle]

Does that make sense?

So angle DCO is 2x

You get why?