help-39

[regal herald]

i decided to say each of the four sections of the beam has length l

so they are both just l away from the pivot yeah

[lucid hazel]

can you please help me with this

i know more birhgtness = less resistance

[pearl pond]

@lucid hazel Has your question been resolved?

[pearl pond]

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[pearl pond]

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[limpid spear]

Is this the right equation

[bitter lodge]

This isnt right

Can you how how you got y = 2x - 100

[limpid spear]

I just re-did it and got y=2x+300

[bitter lodge]

✅

[limpid spear]

.close

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[orchid dragon]

i dont understand the question, they all seem like simple circuits in some way and not simple paths

[orchid dragon]

i figured it out thank you

.close

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[mystic bison]

yucky

[gritty cipher]

im not sure how to solve this indefinite integral

yeah lol

[boreal marlin]

what substitutions have you tried

the obvious one works

[gritty cipher]

i tried u = 2x^2 + 1

[boreal marlin]

yes that should work

[gritty cipher]

really?

[prisma elk]

yes

[boreal marlin]

yes indeed split the numerator in x^2 and x

and use $x$ for changing the dx to du and $x^2=\frac{u-1}{2}$

[jolly parrot]

VincentBH

[gritty cipher]

so i got $\frac{u-1}{8sqrt(u)}*du$

[jolly parrot]

BlackWizard

[gritty cipher]

??

[prisma elk]

yes

split it

1/8 * sqrt(u)

[versed mica]

\sqrt{u}

[prisma elk]

• 1/8 1/sqrt(u)

[fierce totem]

You can just sub the entire √(2x²+1)=u

[boreal marlin]

I think they mean $$\int\frac{u-1}{8\sqrt{u}}:\mathrm{d}u$$

[jolly parrot]

VincentBH

[gritty cipher]

ya

[prisma elk]

yes

u split

u/sqrt(u)

1/sqrt(u)

do ur basic integration

[gritty cipher]

yeah

[prisma elk]

ok then?

do it

then come back

[gritty cipher]

pretty sure i got it

yo chill out im doin it

xd

[prisma elk]

oke do it then come back

[gritty cipher]

i figured it out thanks

.close

[pearl pond]

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[past jungle]

how do i solve part (c)? i know that geometric series goes like a/1-r but i dont know how it goes here...

[cobalt hinge]

Which FRQ year is this from?

[past jungle]

2014

ap calc bc

[cobalt hinge]

Also, note that we have $\sum_{n=1}^{\infty}(-1)^{n-1}2^nx^{n-1}$.

[jolly parrot]

;(

[cobalt hinge]

What would be the first term of this sequence?

[cobalt hinge]

Time to speedrun 😄

[past jungle]

2^n?

[versed mica]

first term is what you get upon letting n = 1

this is how summations work

[cobalt hinge]

Not necessarily. Do you know how indices work?

[past jungle]

so first term is 2 and the common term r^n is (x-1)?

[versed mica]

are there other parts that are raised to the power of n?

[past jungle]

i dont know how to identify a and r here ive only done it before in standard a*r^n form

[versed mica]

^

tell me if you see anything else with an exponent with n in it

[past jungle]

just n or n+1 and n-1 counts too?

if just n then 2^n?

[versed mica]

x^{a+b} = x^a * x^b

[past jungle]

let me try

[versed mica]

@past jungle how did you do

[past jungle]

i have no idea 😭 😭 i tried reviewing this geometric series thing but kinda failed

[versed mica]

$\sum_{n=1}^{\infty}(-1)^n (-1)(2)^n(x-1)^n(x-1)^{-1}$

[jolly parrot]

knief

[versed mica]

now take all the terms with ^n

and you can combine them into a single ()

since a^n * b^n = (ab)^n

$\sum_{n=1}^{\infty}(-1(2)(x-1))^n(-1)(x-1)^{-1}$

[jolly parrot]

knief

[versed mica]

now just multiply together the terms inside

you get -2(x-1)

[past jungle]

ok let me process this

[versed mica]

if you have a question just ask

@past jungle

[past jungle]

ok

let me write this down

wait how do i identify a and r?

[versed mica]

a is the first term

[past jungle]

yeah so 2?

[versed mica]

r is the ratio or the term being raised to the power of n

[past jungle]

i dont get why r is -2(x-1) cause i thought it was (x-1)

[versed mica]

^

because we can rewrite it like this

do you understand what the ratio really is telling you?

[past jungle]

oh we can open the parenthesis

[past jungle]

ratio is like the repeated term?

[versed mica]

if r is the ratio then it means that any term is r times the previous term

[past jungle]

ohhhh

it makes sense

[versed mica]

since we’re summing over n we notice that with each successive term there’s an additional factor of -1, 2, and (x-1)

if you need to just write out some terms

this is because (-1), 2, and (x-1) all contain n in their exponents

so increasing n by one or just observing the next term multiplies by an additional factor of each of those

[past jungle]

i think i didnt understood this series thing this whole time

need to review this asap

[versed mica]

all calculus struggles are from holes in fundamentals

[past jungle]

so its like we used the general term of f' to write down the geometric series and integrate it make it f? but i dont get that last line in the solution

what is it talking about

[past jungle]

fr

[versed mica]

the |x-1| < 1/2 thing?

didn’t you find this from part a?

[past jungle]

yeah the radius

[versed mica]

so what’s confusing

[past jungle]

f(x) =ln|2x-1| for |x-1|<1/2 what does this mean whats f(x) for?

the quesiton also said determine f for |x-1|<R and are we proving something...?

[versed mica]

they’ve defined what f(x) is equal to then specified a constraint on what x can be which you found from part a

because the series doesn’t converge without that restriction

and the series equals f if it does converge

[past jungle]

ohhh i get it

the interval of convergence?

[versed mica]

wdym

[past jungle]

i dont know what im saying

[versed mica]

lmao

[past jungle]

so its like we found f' which converges at the |x-1|<1/2 and we're gonna use this to find f which converges when |x-1|<1/2

[versed mica]

you found f converges for that

[past jungle]

yeah

[versed mica]

and then you used f’ because it was a geometric series and we know how to convert a geometric series into a function which we can then integrate to find f but we still have the condition that it converges for what we had before

[past jungle]

so the last line is like stating that?

[versed mica]

yes it gave the explicit function rather than a series and it gave you the region for which it converges

[past jungle]

i think i kinda get it

[versed mica]

the taylor series outside of that region doesn’t converge to ln(yadda yadda)

but f = ln(yadda yadda) so they just specified that the taylor series for f = f in that region

[past jungle]

ok

thanks so much for helping this was a long one 🥹

.close

[pearl pond]

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[versed mica]

you’re welcome

have a nice night

[pearl pond]

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[pallid crane]

question 4

[pallid crane]

How do I set up the integral?

[bitter lodge]

You are given the points of intersections

Oh wait

Ok i recommend integrating in terms of y

[pallid crane]

yes, what do they tell me?

[pallid crane]

?

do you mean the intervals should be of the y axis instead of x

actually i think it makes sense

[bitter lodge]

If you rotate the graph 90 degrees, you'll notice its WAY easier to integrate

And theyve even given you a function x in terms of y

[pallid crane]

yeah so I subtract, x+y^2-4y by the other function

[pallid crane]

ok so that from 0 to 3 of the y-axis

alright thanks

.close

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[bitter lodge]

looking for someone to check my work

[bitter lodge]

(#56)

[dense jasper]

,w magnitude and direction of (4+sqrt 3, 1)

[jolly parrot]

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[dense jasper]

,w 1+(4+sqrt3)^2 = 20+8sqrt3

[jolly parrot]

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[dense jasper]

,w arctan(\frac{1}{4+sqrt3})

[dawn sand]

WA knows what its doing with this output

[jolly parrot]

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[bitter lodge]

Are you the guy 798 talks about

[bitter lodge]

Thanks!

[cobalt hinge]

You speedrun that 😭

[bitter lodge]

Ikr

Fingers going like magic

.close

[pearl pond]

Channel closed

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[dense jasper]

.reopen

[pearl pond]

✅

[dense jasper]

you may not have done the right question lol

[bitter lodge]

:(

[dense jasper]

it wants the angles that u and v make individually

not the sum

that being said they give it to you so

idk

[bitter lodge]

But doesnt it give me the angles

[dense jasper]

yeah

that's why I'm like ehhh...

[bitter lodge]

I interpret it as "the angles that (u and v) make with the positive x-axis"

[dense jasper]

🤷‍♂️

all ik is that what you did is correct

.close

[pearl pond]

Channel closed

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[bitter lodge]

Wait

[dense jasper]

.reopen

[pearl pond]

✅

[bitter lodge]

Are you the guy who tortures 798 with all those questions

[dense jasper]

given that idk who 798 is

imma say no

[bitter lodge]

@graceful lava this guy

[dense jasper]

oh

hello that guy

[bitter lodge]

Nah hes not here rnt

[dense jasper]

I could be

[bitter lodge]

But that problem with like 10 diff shapes

[dense jasper]

depends what you mean

[bitter lodge]

It was hard :(

[dense jasper]

uh

what r u talking abt

[bitter lodge]

Big circle + trapezoid in the top left

[dense jasper]

oh

that shit

[bitter lodge]

Lol

Anyways nice to meet you

.close

[pearl pond]

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[pearl pond]

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[placid surge]

Can anyone explain the rules to simplify this in a simple way

[pearl pond]

@placid surge Has your question been resolved?

[placid surge]

I look at this problem and I’m not sure where to start

[agile valley]

what do you mean by simplify? solving for f(x)?

[placid surge]

Even a name of this type of problem and I could try and watch videos on it

Yeah…. I would say solving but another disc member said it’s not solving for f(x) its just implying the expression

[compact ridge]

$\frac{x}{x - 1} - \frac{4(x - 1)}{x - 1}$?

[jolly parrot]

south

[compact ridge]

not sure what you want to do

[placid surge]

Put it in simplest form

[agile valley]

f(x) - 4 is on the left

[compact ridge]

I think they have that f(x) = x/(x - 1)

it's definitely not an equation to solve

[placid surge]

The answer shows to be 4-3x/x-1

But I have no idea how to go about getting that

[compact ridge]

ah okay so I understand now

[compact ridge]

notice how both fractions have the same denominator

so you have $\frac{x - 4(x- 1)}{x - 1}$

[jolly parrot]

south

[placid surge]

Well you don’t start with that

[compact ridge]

I know, I didn't assume the answer

[placid surge]

Ok

What rules allows you to

Go from (x/x-1) - 4 to that

That is my question

[compact ridge]

https://www.youtube.com/shorts/9NKjzwdd7Ng

here's another example

[compact ridge]

you can write 4 as $\frac{4(x-1)}{x-1}$

[jolly parrot]

south

[placid surge]

Where is the -1 coming from and why

I just don’t get it

[compact ridge]

you need a common denominator

if we work with just numbers for example

let's say you want to write 5/7 + 4 as an improper fraction

we would start off with 5/7 + 28/7

4 = 4 * 7/7 = 28/7

[placid surge]

But there is no other fraction

The video you sent me makes sense bc there are two fractions

And the goal is to add them

[compact ridge]

the trick is to write 4 as a fraction

[placid surge]

That’s just -4/1

[compact ridge]

do you understand how this example works first

we have the same denominator for both fractions, cause we made it that way

[placid surge]

Yeah

I understand finding a common denominator

[compact ridge]

yes, so it's the exact same for the original question

the denominator of x/(x - 1) is x - 1

[placid surge]

Yeah ik

Why are we getting rid of the - part of the 4

[compact ridge]

where?

[placid surge]

F(x) -4

[compact ridge]

ah, so like if you have say 32/7 - 28/7, that is just (32 - 28)/7

[placid surge]

Ok

So even if I do 4(x-1)/x-1

I just get 4x-4/x-1

[compact ridge]

yep

[placid surge]

Sooo what am I supposed to do next

[compact ridge]

you're nearly there then

$\frac{x}{x - 1} - \frac{4x - 4}{x - 1} = \frac{x - (4x - 4)}{x - 1}$

[jolly parrot]

south

[compact ridge]

the brackets are super important, otherwise you'll mess it up

[placid surge]

Where is x- coming from

[compact ridge]

it's like how 32/7 - 28/7 = (32 - 28)/7

[placid surge]

I don’t see it

Loosing me with that

All I see is numbers coming from nowhere

When numbers come from nowhere there needs to be a goal

I see no goal

Like it makes sense to me when I need to make an expression positive so I divide it by -1 or something like that

[compact ridge]

the goal here is to write the entire expression as one fraction

ah, do you accept how $-\frac{4x - 4}{x - 1} = \frac{-(4x - 4)}{x - 1} = \frac{-4x + 4}{x - 1}$?

[jolly parrot]

south

[compact ridge]

would that make it easier

[placid surge]

Yeah

That makes sense

Distribution of the minus sign

[compact ridge]

then you have $\frac{x}{x - 1} + \left(-\frac{4x - 4}{x - 1} \right)$

[jolly parrot]

south

[compact ridge]

now cause we simplified that thing already

we have $\frac{x}{x - 1} + \frac{-4x + 4}{x - 1}$

[jolly parrot]

south

[placid surge]

Should I have put f(x)-4 over x-1?

Is that what I’m missing

[compact ridge]

from what you told me, no, I think you understand that step

I think you just need more time to process subtracting algebraic fractions

yeah it is easy to mix up the brackets

[compact ridge]

so I would recommend you do something like this for now

convert the subtraction to addition by distributing the minus sign

to the other fraction

[placid surge]

Can you rewrite the equation you want me to practice and I’ll do it

[compact ridge]

https://www.madasmaths.com/archive/maths_booklets/basic_topics/various/algebraic_fractions.pdf

here's a whole workbook, start from page 11

[plush bramble]

pro googling

[compact ridge]

no, I just know of MadasMaths

they're really good for grade school stuff + first-year uni

[plush bramble]

i like it!

[placid surge]

Thank you I’ll go through it

[pearl pond]

@placid surge Has your question been resolved?

[pearl pond]

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[knotty apex]

This picture was translated into English by a translator and may not be accurate.Can someone please explain to me how to do this problem, really not, please

[pearl pond]

@knotty apex Has your question been resolved?

[latent quail]

先決定每所大學的報名人數是多少

這部分你會算嗎？

[knotty apex]

会的

[latent quail]

你算出來的組合數是多少？

如果把四位學生當成同一種物品

換個問法，四顆蘋果丟三個不同的袋子有幾種丟法？

每個袋子一定要有蘋果

[knotty apex]

组合数是3，对吗？今天刚讲排列组合，没听懂😞

[latent quail]

對，就是三種

接下來，先選出三位同學把他們安置在那三間學校

多出來的那位下一步再討論

請問從四位同學選出三位的組合數是多少？

[knotty apex]

3种，然后相乘？

3x3

[latent quail]

對

接下來

請問把選出的三位同學排列在三間大學的排列數是多少？

[knotty apex]

这个什么意思？脑子有点转不过来

[latent quail]

有點像是三個人排成一列，有幾種排法

ABC BCA CAB…

[slate mirage]

Try making him understand using 2 students first

once they understand it, is multiples of power they will understand 3^3 and 3^4

@knotty apex

這是機器翻譯的。 請原諒。

---

讓我們嘗試僅與一名學生一起觀看。 如果 1 位學生可以選擇 3 所大學，那麼他只有 3 個選擇。

---

現在我們先忽略學生1，另一個學生「2」對大學有相同的3個選擇。 所以他有3個選擇。

[knotty apex]

OK，my English is not well ,so I have spend some understanding your word

[slate mirage]

---

因為這個例子允許重複，現在還有其他限制，我們可以把學生 1 和學生 2 結合起來

[knotty apex]

ok,next?

I know this

[slate mirage]

---

我們將學生1 的選擇與學生2 的選擇相乘。 。 您將看到我們有 9 種不同版本的圖紙。

---

現在我們知道 2 位學生和 3 位大學給我們 3²，那麼 4 位學生給我們 3⁴ = 81

[latent quail]

[slate mirage]

並且沒有其他限制，*

sorry I mistype it

@knotty apex

[knotty apex]

ok,I understand

[slate mirage]

我們只能這樣做，因為我們允許重複且不受限制。 這僅針對排列組合中遇到的問題的 1/4

這也是您將看到的 4 個中最簡單的一個。 你不會經常看到這種情況。

希望我能幫助您並且您能理解我。 如果您的問題已解決，請輸入“.close”

If you are in China, good afternoon! 👋

[knotty apex]

ok ,thank you very much .CLOSE

[latent quail]

.close

[pearl pond]

Channel closed

Closed by @latent quail

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[knotty apex]

.close

[slate mirage]

👍

[pearl pond]

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[high garnet]

Can someone correct my work ?

I think I Made a mistake somewhere

My answer is not equal to 140/3

Which it should

[plush moss]

Theres a better way to do this

The ratio of the areas on a similar figure is the same as the ratios of a length ^ 2

[high garnet]

what is a and b then ?

is it the high ?

[plush moss]

Any corresponding length on the figure

[high garnet]

ohh i see

can you help to compute it ?

i dont have any idea

[plush moss]

Uh I'll help you draw a sketch

Zoom in to that pyramid figure

The 3d one

[high garnet]

thanksss

[plush moss]

So you've drawn a smaller pyramid in this larger pyramid yes?

Let's call that smaller pyramid's height x

So what is the length ratio of the smaller pyramid to the bigger one

@high garnet

Express the ratio in terms of x

[high garnet]

(b/5)^3?

@plush moss

[plush moss]

What's b?

@high garnet

[high garnet]

unknown ??

[plush moss]

The unknown height?

Welp I think you got it

I was just asking for the length ratio

And that's b/5

Then the surface area ratio is just (b/5)^2

[high garnet]

ohh really ?

[high garnet]

okay what is the next step then ?

[plush moss]

Alright so now we make an expression for the base of a similar pyramid in the bigger pyramid

The original base has an area of 7*4=28

So A=(b/5)^2*28

So now you can integrate A=(b/5)^2*28 from 0 to 5

To get the volume

Yipee

[high garnet]

waittt

can you help me to draw it by giving a hint or something ?

cuz i still dont get it how it can make a pyramid

[pearl pond]

@high garnet Has your question been resolved?

[high garnet]

@plush moss

[pearl pond]

@high garnet Has your question been resolved?

[plush moss]

Uh tbh I think I have a Yt video that might help

https://youtu.be/WUvTyaaNkzM?si=qagcKuloPl8cGvdu
Watch from 1:33 to 6:47

[rich jolt]

this definitely wont help him

[plush moss]

Idk I don't have something I can draw on rn😭

[rich jolt]

calculate area of one side of the tetrahedron

and the base

pretty easy

[plush moss]

You have to use integration

The only formula you can use is for the area of a rectangle

[rich jolt]

yeah

still doable

hol up not a tetrahedron ithink

still doable

[plush moss]

with integration or some other method

[plush moss]

In the video he finds the formula for the area of a circle by integrating it's circumference

So it's kinda applicable here

In visualising what to integrate

[plush moss]

@rich jolt how does this work without integration though

[pearl pond]

@high garnet Has your question been resolved?

[pearl pond]

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[spice grail]

Kinda need help to understand some sht

[spice grail]

I need someone to explain how we solve this to me :

If X follows a normal law with parameters ( -9 ; 4² ) then P(X inforequal 1.28 ) equals to :

A. 0,995
B. 0,950

C. 0,975
D. 0,990

i'm pretty sure -9 is the center

Pls ping me if you have any answers

we also have this but I kinda want to understand the reasoning behind this

I also a few tons other questions remaining if you have time to waste 🥲

[unborn abyss]

we need to know how many standard deviations away from the mean is 1.28
1.28 - (-9) = 10.28
10.28 / 4 = 2.57

tu dois utiliser un tableau

[spice grail]

I'm gonna guess I have to use this hence why it's 0.95

[unborn abyss]

yeah that's the same info as the table just sideways

or like. just a few points rather than all of them :D

[spice grail]

yeah I get it so for this it's just reading the stupid sht ig

[unborn abyss]

yeah scaling and then looking it up

[spice grail]

okay yeah so I need to do the steps before

just as you did

so like (a,b) and P(x<= c )
zb = c - a
zb/b = za

and then reading

@unborn abyssalso I've seen you typed smthing in french, are you french ? Cuz me too so I could just print the screenshots here lmao

[unborn abyss]

je ne suis française mais je peux lire

[spice grail]

okay good cuz I can understand english it's just that I don't have the specific vocabulary in mind for those

I'm pretty sure here you have to do 2² / n since it's independant which gives -5 1/2 but yet again I know the result not the method and why it applies

which isn't rigorous

[spice grail]

if you want I can try to translate these it shouldn't be too hard

[unborn abyss]

i think i understand, just trying to remember how sample standard deviation works :D

[spice grail]

Ahaha thanks for helping and also good luck, we had to do those in excel for exams but I miserably failed with data bases so now I have to do everything again

[unborn abyss]

oui, est 2^2 / 8 = {-5, 1/2}

ummmm the idea here is that imagine you took 1000 samples and averaged them. you are almost certainly going to get a moyenne of -5

[spice grail]

okay thanks this one shouldn't be hard

[unborn abyss]

but if you only take 3 samples there's a good chance you get a mean like -1 or -7

[spice grail]

oh yeah I see

this one seems tought from what I've seen

[unborn abyss]

what is lambda here?

[spice grail]

I wouldn't know

it only says this

but I reckon it's for p

[pearl pond]

@spice grail Has your question been resolved?

[pearl pond]

@spice grail Has your question been resolved?

[pearl pond]

@spice grail Has your question been resolved?

[pearl pond]

@spice grail Has your question been resolved?

[spice grail]

.close

[pearl pond]

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[velvet lion]

hello

[velvet lion]

is my proof right?

this is a theorem in class not hw or assignment btw

this is supposed to be proof of i)

[pearl pond]

@velvet lion Has your question been resolved?

[rough forge]

Doesn't the theorem assume that f is invertible, hence bijective? So your counter example wouldn't work?

[velvet lion]

Which is exactly why I’m so confused

I don’t know what is happening

I didn’t right the counter example my professor did

And then told us to prove the theorem

[rough forge]

what

[velvet lion]

I am lost confused

If f is invertible, does my proof hold?

[rough forge]

maybe you are suppsoed to prove it for functions that are invertible

[velvet lion]

😭😭 probably yeah

[rough forge]

Well I was confused about your proof at first because you just stated if x is in A then you claimed that there exists a y of which I was confused

But indeed, if f is bijective then yes there must exist some kind of y for x such that y = f(x)

due to bijectivity

[velvet lion]

Oh :((

I wanted to show if x was in A Then x is also in f-1(f(A))

So first I assume there is an x in A

since I was also told there was a function f:X->Y (this is assumed)

Then there has to be some y that is in f(A) right?

[rough forge]

that is a bijection

and then yes

[velvet lion]

Um

If A is a subset of f-1(f(A)), why us it 1 to 1 and onto

Cuz they’re not “equal”

[rough forge]

hmm i dont think i understand

my point was that i think f has to be well defined

[velvet lion]

I’m still having trouble understanding so I think I’ll make a run to my prof rn

But I appreciate you trying c: have a good day

[rough forge]

I think I was actually wrong

[velvet lion]

Even if you are it’s okay

Would you like me to share my solution with you after I find out? (If ur interested?)

[rough forge]

yes

[rough forge]

also did you write that it was a counter example?

cause it rather seems like an example

f(A) should be also {0,1} and so f^-1({0,1}) = {-1, 0, 1} and A is a subset of that

a strict subset

[velvet lion]

Mmm I wrote that little side because I heard the professor saying it

I put a question mark since I wasn’t sure

I’ll tag u back later when I find solution

[pearl pond]

@velvet lion Has your question been resolved?

[pearl pond]

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[midnight flicker]

In a regular quadrangular pyramid (a square at the base), the side edge forms an angle beta with the side of the base. Determine the side surface of the pyramid if the radius of the circle inscribed in the side face is equal to r

[pearl pond]

@midnight flicker Has your question been resolved?

[pearl pond]

@midnight flicker Has your question been resolved?

[velvet lion]

I found out my proof was wrong

I was wrong because I assumed injectivity in the proof!

Let a0 in A. We know f: X-> Y. Then there exists y0 in f(A) such that f(a0)=y0. This is just by definition of image. Then suppose f(A) = B. Since y0=f(a0), by definition of preimage, a0 is in f^(-1)(B). So a0 in f^{-1}(f(A)) is true

[rough forge]

Then there exists c in f^-1(f(A)) right

[rough forge]

yes looks good

[pearl pond]

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[velvet lion]

yup x= c was assuming injectivity

[velvet lion]

:D whoo

[rough forge]

are you studying pure maths?

[velvet lion]

oh heelll naww

im a computer science student!

[rough forge]

ewwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

tomorrow's my datastructure exam

[velvet lion]

oohhh

[rough forge]

y'all abusing too much of my beloved mathematical language

[velvet lion]

good luck :D

LMAO

i agree

[rough forge]

thank :D

[velvet lion]

cs profs abuse math notation

nothing is rigorous in the cs slides

[rough forge]

only rigorous is the professor when they grade your shit 😄

[spare lark]

My grade written in scientific notation

[pearl pond]

@velvet lion Has your question been resolved?

[velvet lion]

LMAO

[pearl pond]

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[plain matrix]

helo

[plain matrix]

not sure how they dividedi by x-2

[leaden wadi]

Do you mean from line 1 to line 2?

[bitter lodge]

Where did they divide by x - 2

[plain matrix]

yea from line 1 to 2

[leaden wadi]

That's just long division.

[plain matrix]

oh cos when i divided i got something else

how would u do the long division?

[bitter lodge]

Can you show what you got

You might have forgotton the -

[plain matrix]

ok i got x^2 - 2x + 1 - 1/x-2

think i did it wrong tho

[plain matrix]

can u show a pic of how to do the long diivision?

[leaden wadi]

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

That eliminates the first two terms.

[plain matrix]

ohh do u just multiply top by x-2?

[bitter lodge]

I feel like this needs to be done with a video

[plain matrix]

oh i see

[bitter lodge]

Polynomial Long Division

[plain matrix]

oh ok i think i dont this before

i will watch the video

thank you

[pearl pond]

@plain matrix Has your question been resolved?

[pearl pond]

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[lapis lynx]

two squares inscribed in a semicircle, how do we find the radius?

[lapis lynx]

36 is the area btw

[winged flicker]

The large square has the center of the circumference as the point that divides one of its sides in half?

[lapis lynx]

yea i drew it bad

[winged flicker]

And does the small square have one of its vertices in a similar situation?

[lapis lynx]

no

[winged flicker]

I meant that, I understand that you have already clarified it for the large square, but does the same thing happen for the small one?

[lapis lynx]

nope

[winged flicker]

Ok

[pearl pond]

@lapis lynx Has your question been resolved?

[pearl pond]

@lapis lynx Has your question been resolved?

[pearl pond]

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[quartz nacelle]

Help needed on setting up the equation 😰

[pulsar lark]

Begin with the diagram

[quartz nacelle]

Is this the diagram?

[pulsar lark]

Looks good

[quartz nacelle]

Yeah idk how to go on from here🥲

[blissful cloak]

you have similar triangles

[fair creek]

the two triangles are similar-

yeah

so just take their ratios of base and height

[quartz nacelle]

This makes a quadratic soooo do I make them equal 0 or go the square root route?

I’m confused 😕

[pearl pond]

@quartz nacelle Has your question been resolved?

[pearl pond]

@quartz nacelle Has your question been resolved?

[pearl pond]

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[full folio]

hello does someone mind helping me with a calc 3 problem?

[full folio]

just need some advice

I was thinking that this is true

Because wouldn’t f(x,y) = 0 make it so that the minimum is 1?

Also since it’s in an absolute value

<@&286206848099549185>

[dusk horizon]

Yes.

[full folio]

thanks!

.close

[pearl pond]

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[next forum]

does this look right?

[odd surge]

!original

[pearl pond]

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.

@next forum Has your question been resolved?

[next forum]

uhhhh one sec

its a multistep problem and all i wanted to know is if the solution to my inequality was right

[odd surge]

shouldn't it be 15x + 25 because admission is a one time thing

idk

wait nvm

[winged flicker]

the answer for the first question is correct

[winged flicker]

For the second problem, understanding that parking is hourly and the admission cost is one-time, then I think the answer would be B.

[next forum]

admission isnt a one time thing, each visit. he wouldnt have to pay for a membership if it were free infinitely after that

each time he visits the museum
that means he pays each time, unless he gets the 110 dollar membership, in which its only a one time payment
thats how i got my inequality (the second question actually came before, discord just loaded the screenshots wrong)

sometimesit reverses the order of my screenshots 😭

[winged flicker]

ok ok, you're right, so in that way your answer would be correct.

[next forum]

sorry for the confusion LOL

and thank you for the help!!

.close

[pearl pond]

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[pearl pond]

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[steel flare]

!status

[pearl pond]

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

[steel flare]

I have my answer

I'd like to get it checked

I got limit does not exist

[rough forge]

ok send

[steel flare]

i didn't complete it

look

the +x and the -x

cancel each other out

ur left with

take the square root outside

now look

you have x^2 + 1

divided by x^3

$\frac{log(x^2+1)}{x^3}$

[jolly parrot]

kronium_

[steel flare]

as limit x --> 0

now look

since theres a 1

x^2 in the log is insignifcant

so log 1 = 0

x^3 also equals 0

now if

x is ...

positive

or negative

x^2 is always positive and goes to 1

[fierce totem]

[steel flare]

x^3 determines if the limit is negative or positive

it either goes to positive or negative infinity

and it that happens

then limit does not exist

since there isn't continuity

thx targetvpn

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[pearl pond]

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[rough forge]

i still dont get how you derived dne

0/0 is indeterminant form

[azure notch]

This is what I got:

Idk if right

The step I don’t know is right is the last one

Where I cancel x instead of apply l hospital again

My reasoning was x is never zero but actually now I think abt it I don’t think that’s right

[rough forge]

yea you simplifed wrong

after lhopital

[rough forge]

the forms looks similar to 1/x

[steel flare]

graph tells a different story

[rough forge]

yea but your argument is weak

[azure notch]

Yeah

[rough forge]

you like derived dne out of nowhere

when 0/0 could be anything

[fierce totem]

0/0 can be anything and in this case it DNE

[rough forge]

yes and i fail to see where you got to the point where it clearly dne

[bitter lodge]

Where did absolute value come from? Or is it because x^2 + 1 > 0

[steel flare]

ok fine

[rough forge]

if you apply here lhopital it gets more obvious why it dne

[azure notch]

This is dne right

[steel flare]

IT DOESN'T MATTER WHETHER OR NOT THERE IS ABSOLUTE VALUE

the rresult does not change

[rough forge]

yes this clearly shows it

[steel flare]

the x there goes to zero

[azure notch]

Yeah mb

[steel flare]

therefore the result goes to positive or negative infinity

[fierce totem]

Probably the simplest solution is to approximate ln(x^2 + 1) = x^2

[steel flare]

ok imma go

[fierce totem]

then the limit is just 1/x

[steel flare]

i got a very important exam tmrw

i gotta study

cya

thx again @fierce totem

[fierce totem]

k. gl

[pearl pond]

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[vale geyser]

Hello! I’m having some difficulties with factorizations. For exercise 8, I’m not sure about my results. I thought that (3x + 1)^2 and (2x − 1)^2 were remarkable identities, so I used them. However, I’m wondering if I actually expanded instead of factoring.
Can anyone help me plz?

[cedar bough]

use difference of squares

[vale geyser]

Okay, I'll try to do it this way. I wasn't sure how to go about it

Thanks

[cedar bough]

yeah in this way it'll be directly in factored form

so you don't need to do much after that

[vale geyser]

Is it better like this🥲 ? Then I simplify?

[cedar bough]

the brackets are a bit off

[vale geyser]

brackets are not necessary?

[cedar bough]

[ (3x+1) - (2x-1) ] * [ (3x+1) + (2x-1) ]

[vale geyser]

Ahh okayy

[cedar bough]

see the difference?

[vale geyser]

Yeah I think

[cedar bough]

yeah you made it look like the middle 2 terms are being multiplied first

that's it

you can continue from here and you'll get a nice factored form

[vale geyser]

Maybe it’s better, or maybe it’s just a disaster

[cedar bough]

5x(x+2) is just done

you needed to factorise so it's done

you need not expand further

and it's correct

[vale geyser]

So I'm stopping at 5x (x + 2)?

[cedar bough]

yeah

and that's your answer

[vale geyser]

Okay, thank you for your helps !

[cedar bough]

you're welcome

[vale geyser]

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[dapper token]

fast question

btw at the end is the basis of these vectors

the first matrix is 5 vectors

and i need to find the basis of them

anyways

on the 3rd matrix

[vital crescent]

what do you mean by that?

[dapper token]

oh

sorry forgot to continue

like

how did he know that he doesnt have to do C2 -> C2-C1

uk what i mean ?

[vital crescent]

on which step?

[dapper token]

like why didnt he remove the one from the 2nd column even tho he cld

top right matrix

3rd matrix

from 3rd to 4th

[vital crescent]

wouldn't C2->C2-C1 make C2 -1,0,-2,-4?

[dapper token]

ok ur right

actually

that answered for another follow up question that i had lmfao

tysm

[vital crescent]

nw

once you get C2 to be the "tallest" column (but shorter than C_1), you can just keep it still

[dapper token]

ahhh

[vital crescent]

just make the top entry 1

cuz you know you can just remove a multiple of C_2, to reduce the rows C_3,C_4,... to some column with height strictly less than C_2

[dapper token]

ye gotcha man

tysm i appreciate it

.close

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[visual relic]

Hello! I'm new here and I have a question because i missed a day in school. I learned in the book that when we have f(x)--->f(x+2), f(x) moves left. But let's say f(x)=x², when we have x=-2 we get f(x)=4 and f(x+2)=0. But when we have x=2 we have f(x)=4 and f(x+2)=16, from 4 to 16, doesn't f(x) move on the right and not on the left? Sorry if my question is not clear but english is not my first language T-T

[midnight haven]

f(x) = x+2

[edgy stone]

that's f(x) -> f(x+4)

not f(x) to f(x+2)

[visual relic]

Ohhh i see😭

[pearl pond]

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[warm current]

images are preferred

[bitter lodge]

Which problem do you need help with?

[midnight haven]

what's the thing for images??? jpeg?

[bitter lodge]

Do you know how to find vertical asymptotes, and how to differentiate them from holes?

[midnight haven]

i think it's -b/2a?

[bitter lodge]

thats something different

First, factor the numerator and denominator

[midnight haven]

How do I factor them?-

[bitter lodge]

-4x + 8

What do you notice that's common

[midnight haven]

that they are both devesible... by 2 and 4

[bitter lodge]

So lets factor out 4

-4x + 8 = 4(-x + 2)

There is nothing more to factor

Now lets work with x^2 + 2x

Can you factor that?

[midnight haven]

nope

[bitter lodge]

Ok, think of the numerator as x * x + 2 * x

[midnight haven]

because there is nothing else to factor?

[midnight haven]

There's nothing else to factor

[bitter lodge]

There is

You can factor out x

[midnight haven]

What-

But Ax^2 + Bx + C

[bitter lodge]

$$x^2 + 2x$$
$$x(x) + 2(x)$$
$$x(x + 2)$$

[jolly parrot]

King Leo

[bitter lodge]

In this case, C = 0

[midnight haven]

but there is + 8?

WAIT

Ohhh

okay

It will be become x + 2?

[bitter lodge]

The + 8 is for the denominator

So our original function can be rewritten:
$$\frac{x(x + 2)}{4(2 - x)}$$

[jolly parrot]

King Leo

[midnight haven]

okay

[bitter lodge]

So do you know what must happen for a vertical asymptote to occur

[midnight haven]

No-

[bitter lodge]

The denominator must be equal to 0

What value(s) of x set the denominator equal to 0?

[midnight haven]

0?

[bitter lodge]

No, because $4(2 - x) = 4(2) = 8$

[jolly parrot]

King Leo

[midnight haven]

replace 4 to 0???

[bitter lodge]

You cant change 4

The only thing you can change is x

[midnight haven]

4(2 - (0)) ?

NO WAIT

4(2 - 2)?

4(0) = 0?

@bitter lodge

[plush bramble]

,calc 4 * 0

[jolly parrot]

Result:

0

[plush bramble]

yes

[midnight haven]

YES

So it's just x(x+2)

[bitter lodge]

No

x = 2 is an asymptote

Because that sets the denominator equal to 0

And when you start approaching a very tiny denominator, your overall y value skyrockets (almost literally) in the positive or negative direction

[midnight haven]

so x = 2 is the vertical asymptote

while the horizontal asymptote is x = 0?

[bitter lodge]

Horizontal asymptotes need to be in terms of y = ...

[midnight haven]

so y = 0?

[bitter lodge]

Technically yes, but not just because thats where the numerator is 0

[midnight haven]

Okay

so I got 2 answers in 1

Vertical : x = 2
Horizontal y = 0

@bitter lodge

[bitter lodge]

Wait

I messed up

[midnight haven]

Wha

[bitter lodge]

Dyk about slant asymptotes?

[midnight haven]

Nope

@bitter lodge

<@&286206848099549185>

<@&286206848099549185>

[plush bramble]

have you seen this before
https://www.mathsisfun.com/algebra/polynomials-division-long.html

this is the method you use to find slant asymptotes

if not this is a good start
https://www.youtube.com/watch?v=RPXMBIFG_W4

[midnight haven]

Nope

[plush bramble]

not sure what methods you learned to find slant/oblique asymptotes then

[plush bramble]

but the function definitely has a slant asymptote. you can tell because the order of the polynomial in the numerator is 1 more than the order of the polynomial in the denominator

[midnight haven]

But it isn't a uh

It's not a Ax^2+Bx+C

Hm

[plush bramble]

synthetic is another method that could be used

[midnight haven]

The answer will be 5x^3 + 6x

[plush bramble]

https://brilliant.org/wiki/synthetic-division/

[midnight haven]

Okay Time to use chatgpt

Because I have no idea any of this

thank you.

.close

[pearl pond]

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[pearl pond]

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[iron sorrel]

how did they find that these were equal when t = pi/6

[open rivet]

,w sin(2/3 * pi)/cos(2/3 * pi)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by .bonk__

[open rivet]

like this

[pearl pond]

@iron sorrel Has your question been resolved?

[pearl pond]

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[daring anchor]

So a derivative is the change in a slope over a distance infinitely approaching 0?

[plush bramble]

"infinitely" is not necessary here

[daring anchor]

And an integral is the total area underneath an a section of the curve?

[daring anchor]

Ok

[plush bramble]

signed* area. shaded area underneath the x-axis counts as negative

,w int 0 to 2pi sin(x) dx

[jolly parrot]

Results provided by WolframAlpha

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Requested by rie.mann

[plush bramble]

,w plot sin(x) for 0 < x < 2pi

[jolly parrot]

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[plush bramble]

The + area above the x-axis exactly cancels the - area below the x-axis for sin(x) between 0 and 2 pi

[daring anchor]

So the integral of this function would equal out to 0?

[plush bramble]

yes

[daring anchor]

So it is from the y value to y = 0?

And it is over span a to b?

$\int_{b}^{a}(x^2dx)$

[versed mica]

\

[pulsar lark]

What are you trying to do

[jolly parrot]

Marshall The Gamer

[pulsar lark]

$\int$

[jolly parrot]

Offie

[pulsar lark]

Latex easy

[daring anchor]

So in:

\begin{align}

\int_{b}^{a}(x^2dx)

\end{align}

a integral calculates the area spanning x of a to x of b?

[jolly parrot]

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

[daring anchor]

Where the area of a specific x is the derivative of y or 3(y^2)?

<@&286206848099549185>

[pulsar lark]

Wdym area of a specific x?

[daring anchor]

Like when you try to find the area underneath a specific cordinate such as (7, 5) you find a derivative by plugging in the y value into 3(y^2) right?

[plush bramble]

use area of rectangle = length * width

[daring anchor]

Ok, so you us a rectangle of length Δx for width and height y?

And you use a derivative for x so it approaches 0

[plush bramble]

what are you trying to calculate

area or slope of something?

[daring anchor]

Just trying to understand derivatives and integrals

And i know that to find an integral you have to understand derivatives so i am just making sure i understand them

@ me if you respond

[plush bramble]

are you watching any videos or reading anything? it's better if you have specific questions about integrals/derivatives

[daring anchor]

I am watching 3b1b fundamentals of calculus

[plush bramble]

that's uhhhh pop math and mroe for getting you interested, not so good for teaching you fundamentals

[daring anchor]

Any recommendations that a 13yr old could understand?

[plush bramble]

no

[daring anchor]

Any general recommendations?

[plush bramble]

learn functions and trigonometry first

[daring anchor]

Function are equations that can be graphed and trig is triangles and finding side lengths and angles off of other given properties

Yeah?

And using exponents applied to x in functions adds curves to slopes in graphs yeah?

Even exponents create a parabola, and odd ones create curves that go from quadrant 3 to quadrant 1 yeah?

@plush bramble

[plush bramble]

Even exponents create a parabola
no

i don't know what you're trying to prove

naming a few random facts from 2 years worth of material doesn't demonstrate anything

[pearl pond]

@daring anchor Has your question been resolved?

[versed mica]

it’s good that you’re excited about learning math but you clearly are rushing to learn things from crash course math videos which are meant to entertain you

if you actually enjoy math then why not just do math at your level then progress to the more advanced things instead of trying to be able to say you know something to try and impress someone

and the 3b1b playlist is a good playlist for gaining an intuition for things but to actually learn things you can’t use those videos as a substitute for a textbook

he mentions that

[daring anchor]

I do not try to impress people, i just get bored with the stuff on my level

[pearl pond]

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[buoyant garnet]

One message removed from a suspended account.

[buoyant garnet]

One message removed from a suspended account.

One message removed from a suspended account.

[golden bramble]

Follow the instructions?

what is the most obvious u-sub you can make?

[buoyant garnet]

One message removed from a suspended account.