help-28

[sacred sparrow]

then form the derivative:

if we want to know where the derivative is negative:

then we can divide by the e-term

rearrange (mult by root and x)

and do a test-value of f.i. x=1000

which is true

since the right side is a strictly growing function

the inequality does hold for all x>=1000

meaning the derivative is negative for x>=1000

meaning a < b < c

since c has the smallest input (x=2014)

and a the largest (x=2016)

@torn jolt

[torn jolt]

oh i see

the way you are going

well yeah

but

[sacred sparrow]

hm if we were to use log rules instead

[torn jolt]

too much work i feel like this is overcomplicating

very

log base 2015

[sacred sparrow]

probably, is this task in context of some number theory class?

[torn jolt]

no

an university admission exam

in 2016

[sacred sparrow]

hm I'd have done the above approach then, I dislike solving these kind of "what is bigger" questions with odd ways otherwise

[torn jolt]

wait let me try logs

[sacred sparrow]

sure you'd first receive

no not =

hm don't see a direct way to compare the two fractions

I'd again do it with functions :D

[torn jolt]

i did it with logs

using pure logic

[sacred sparrow]

k with which steps

[torn jolt]

its c > a > b

[sacred sparrow]

uh no

I've already proven otherwise above

[torn jolt]

a = sqrt(2014) log_2015(2016) , b = sqrt(2015) , c = sqrt(2016) log_2015(2014)

[sacred sparrow]

ys, continue

🐧

[torn jolt]

now log_2015(2016) is a bit bigger than log_2015(2014)

just a bit

[sacred sparrow]

since?

[torn jolt]

awit

ok yeah

2016 > 2014

its like you compare

log_2(3) with log_2(8)

log_2(8) >

now because of that

[sacred sparrow]

ah nvm you had same base

ys sure

[torn jolt]

that means a > c

[sacred sparrow]

nop

[torn jolt]

well not really

but

close to

wait

[sacred sparrow]

:D

[torn jolt]

what if i note

okaty log

c > a

because

note sqrt(2016) = a

wait anything else

cuz a take

lets say

s

s = sqrt(2016)

then you get

a= s - sqrt(2) log_2015(2016)

b = s - sqrt(1)

c = s log_2015(2014)

[sacred sparrow]

a = s-sqrt(2)*log_2015(2016)?

[torn jolt]

ye

or does that not true

with the sqrt(2)

should i put 2

[sacred sparrow]

sqrt(2016) is not sqrt(2014)+sqrt(2)

[torn jolt]

ok yea

sure

[sacred sparrow]

which makes this equation invalid sry :/

tbf I don't see a direct way to compare them either

[torn jolt]

ok another way

since 2016 > 2015 > 2014

sqrt(2014) < sqrt(2015) < sqrt(2016)

but we are dealing with huge numbers

so

a > b > c

because 2016 > 2015 > 2014

and exponent small change wont make a very much difference

[sacred sparrow]

but the solution is a<b<c

[torn jolt]

bro

ufffffff

this problem is like what

hardvard admission

i dont even know

[sacred sparrow]

is it from harvard?

[torn jolt]

no

LOL

[sacred sparrow]

k, I think the two analytical approaches I mentioned before are probably the way to go

or, if there's a quicker way

then it's likely a number theory proof

that uses either estimates or the properties of sqrt & log

[torn jolt]

im curious how my professor would solve this

[sacred sparrow]

prob analytically

bc it's bulletproof

doing it with log / sqrt rules is kind of hit or miss instead

[torn jolt]

what iffffff

WHAT IFFFFFF

[sacred sparrow]

you might need to try multiple approaches until something sticks

[torn jolt]

ok wait i think i am einstein

[sacred sparrow]

k experiment as you wish

[torn jolt]

yeah i

i got

no

[sacred sparrow]

😄

[torn jolt]

ok so

let me look at soemthing

I CRACKED IT

its c > b > a

look

i might be a geniu

s

so

i found an easier way

wanna see it?

[sacred sparrow]

sure go ahead

[torn jolt]

so i took natural logs of each numbers

you get a = sqrt(2014) ln (2016)

b = sqrt(2015) ln(2015)

c = sqrt(2016) ln(2014)

so far so good

but

that just ended the problem

because

sqrt(x) grows bigger than ln(x)

hence

c > b > a

[sacred sparrow]

that doesn't prove c > b > a

[torn jolt]

it does

?????

[sacred sparrow]

you'd instead need to show that sqrt(x) grows faster than ln(4030-x) decreases

which is the second approach I mentioned before

[torn jolt]

no

this way is the easier way

and its correct

[sacred sparrow]

nop

[torn jolt]

yes

im going to present this to my professor and see what he says

[sacred sparrow]

take sqrt(5) * ln(6)

sqrt(6) * ln(5)

your above logic states the latter is larger

which is false

[sacred sparrow]

he'll chuckle and give you a pat on the shoulder :P

[torn jolt]

:(

[sacred sparrow]

nw it's likely some simple rule that allows for comparison

that we didn't see before

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

uh

[full forum]

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[pliant vigil]

<@&286206848099549185>

[spiral vigil]

do you see a pattern?

[pliant vigil]

yea

its a differemce of three

[spiral vigil]

can you extend it?

make a table

[pliant vigil]

24,21,18,15,12

so on

[spiral vigil]

those are just numbers (and one of them is wrong)

what do the numbers mean?

[pliant vigil]

distance

girl i get it

these values are in arithmetic progression

[spiral vigil]

okay, so what's your question asking?

[pliant vigil]

distance in 8t second

[pliant vigil]

so i gotta find the 8t number in this arithmetic sequence!

24,21,18,15,12,9,6,3..

3 is the answer?!

@spiral vigil

[spiral vigil]

the question is a little bit confusing because it might mean the total distance in 8 seconds, or it might mean the distance in the eighth second

[pliant vigil]

no i asked my teacher

e said the distance covered in te 8t second

[spiral vigil]

ok

[pliant vigil]

but but but

i was told to solve this problem usin some formula

of arithmetic progression

dyk?

??????????????????????????????????//

**did you mind my attitude the last time we met? *

<@&286206848099549185>

[torn jolt]

Find 8th term of the ap

Given first term is 24 and common difference is -3

[pliant vigil]

??

[full forum]

@pliant vigil Has your question been resolved?

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[real osprey]

How can I simplify this boolean algebra expression?

[rugged void]

what the hell is boolean algebra

oh i see

[real osprey]

It's computer science but relies heavily on maths

[rugged void]

just googled it

thought it might b related to vectors, so i was gonna help but now i can't

sorry

:(

[real osprey]

npp

[pliant vigil]

This is a CS question tho

[real osprey]

i understand but maybe there can be someone who would be able to help. I'll close this down

[sacred sparrow]

sure just use distributive and de morgans law

which jet you factor out all terms in KNF

[real osprey]

When it comes to these I struggle to think about what to start with first

[sacred sparrow]

k, let's start on the right

constants first

(B+1)

is the same as (1)

[real osprey]

yes

[sacred sparrow]

k, and then notA * 1 is the same as notA

[real osprey]

yes

[sacred sparrow]

I'll use latex to make it clearer

what I refer to

ok now we have this term:

on the right

what does it simplify to?

[real osprey]

A

[sacred sparrow]

careful

A+0 = A

A+notA = ?

[real osprey]

1

[sacred sparrow]

exactly!

[real osprey]

ahh right

[real osprey]

that side is sorted

[sacred sparrow]

and C*1 = C

yop

now let's look on the left

now comes morgans law into play

which states the following:

not(A+B) = notA * notB

[real osprey]

yes

[sacred sparrow]

not(A*B) = notA + notB

it's btw for longer sequences also true

e.g.

not(A+B+C+D) = notA*notB*notC*notD

which we'll need to use here

[real osprey]

right okay

[sacred sparrow]

I'll group it in brackets:

[pliant vigil]

can i ask wat class are you studying?

[sacred sparrow]

now what comes out if you apply morgans law now?

[pliant vigil]

i want to know te level at which they teach this boolean algebra

[sacred sparrow]

either cs uni or school

rarely discrete mathematics

[real osprey]

A.NOT(B * C).NOT(B * notC)

[sacred sparrow]

Yup! 🦇

[real osprey]

almost there 😄

[sacred sparrow]

ok so now we're here

hmm

can you simplify the not(B*notC)?

this

[real osprey]

give me a sec

[pliant vigil]

bro cuz i also learnt boolean algebra in tenth grade but it wasnt this complex

[real osprey]

notB+c

I think

[sacred sparrow]

kk, there's more puzzles to it :)

[sacred sparrow]

🐛 flawless

[real osprey]

wo i should be more confident 🙂

[sacred sparrow]

ok what about not(B*C)

[real osprey]

notB + notC

[sacred sparrow]

oui

ok, now we have used de morgans law as much as possible

lastly we only need to use distributive law

X*(Y+Z) = X*Y+X*Z

[real osprey]

yes

[sacred sparrow]

k, so if we multiply A with the first bracket we get?

[real osprey]

A.notB + A.notC

[sacred sparrow]

now the slightly tedious step

of these two brackets

[real osprey]

ah expanding double brackets

[sacred sparrow]

distributive law still works the same

(X+Y)*(V+W) = XV+XW+YV+YW

it just becomes a bit longer

[real osprey]

yup

[sacred sparrow]

it's like distributive law for numbers

[real osprey]

so it becomes AnotB.notB + AnotB.C + anotC.notB + AnotC.C

[sacred sparrow]

yes!

[real osprey]

notB.notB is 1

so A.1 = A

1*

[sacred sparrow]

careful, notB*notB = notB

[real osprey]

Ah so that's what I need to remember

Is there maybe a resource that has all these rules including the one you just mentioned?

[sacred sparrow]

you can also insert some logical statement into the variable if you're unsure

not(it rains) AND not(it rains) = not(it rains)

[sacred sparrow]

sure I can look it up in a sec

[real osprey]

great

[sacred sparrow]

ok so we're here

the fourth one should be clear too

how to simplify

[real osprey]

AnotB + AnotBC + AnotCnotB + A.0?

[sacred sparrow]

yes!

well and +C at the end

[real osprey]

ofcourse

[sacred sparrow]

and what is A.0?

[real osprey]

0

[sacred sparrow]

yop!

now +0 is irrelevant, so we can remove it

[real osprey]

Factor out A?

[sacred sparrow]

we never need to factor out btw

[real osprey]

ah okay

[sacred sparrow]

currently it's called the KNF form

[real osprey]

I see

[sacred sparrow]

where everything is separated by +

() + () + () + ...

that's the form you usually want

[real osprey]

okay

[sacred sparrow]

which is the most simplified

now, we're almost done, there's only one last rule to be applied

which tends to be one of the trickier ones for most who get into boolean algebra

let's look at terms 2 and 4

notice that the 2nd term includes the 4th

meaning the C is inside AnotBC

[real osprey]

right

[sacred sparrow]

in that case, the LONGER term can be removed

why is that? because if C = True

then C will obviously be True

however the longer one may not be

but since we're only interested if one of them evaluates to True

only the C term will be relevant

[real osprey]

By includes, are we disregarding the not?

[sacred sparrow]

if that was a bit confusing dw, it'll become easier over time

[sacred sparrow]

we're regarding it

e.g. AnotB does not contain B.

[real osprey]

yes

[sacred sparrow]

ok so by removing the 2nd term we get:

now, can you see another term where this applies?

[real osprey]

The last 2

[sacred sparrow]

hm in what way

[real osprey]

If C is true then it is true

The single C

[sacred sparrow]

almost, because AnotCnotB does not contain C

so here the negation caught you :D

[real osprey]

ahh yes

[sacred sparrow]

however, AnotB is included in AnotCnotB

meaning we can throw AnotCnotB into the volcano

[real osprey]

yup

[sacred sparrow]

🦇

[real osprey]

And thats the answer

[sacred sparrow]

yop!

[real osprey]

thank you so much you have been the biggest help

[sacred sparrow]

absolutely np

the above steps are btw generally applicable

[real osprey]

Is there a way to save these messages or will I need to screen record?

[sacred sparrow]

- simplify constants
- use morgans law as much as possible
- use distributive law as much as possible to get KNF
- simplify KNF using INCLUDED terms
Done!

[real osprey]

Thank you 😄

[sacred sparrow]

hm I can resend them in DMs if you want

[real osprey]

Ah they are saved at the pins

[sacred sparrow]

ofc they remain on the server

but you may not find them anymore

[real osprey]

That is true

[sacred sparrow]

or pins yea

I think here is a fairly comprehensive list

should include most relevant ones yeah

[real osprey]

Thank you

[sacred sparrow]

btw is the "INCLUDED" rule somewhat clear?

bc it's probably one of the hardest when beginning

[real osprey]

Yes, but I will have a more indepth look too

[sacred sparrow]

wish thee good luck, come back any time when stuck 🐦

[real osprey]

Thank you!

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[carmine mango]

not sure if the function is independent or not

<@&286206848099549185>

[rugged void]

!15m

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[carmine mango]

my bad

[full forum]

@carmine mango Has your question been resolved?

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[spark sluice]

would i be right to think that this is false true false?

[sacred sparrow]

nop

[1] and [3] are correct

and essentially have the same meaning

they'd have an infimum / supremum, but not a minimum / maximum

however for [2]

let me give you a subset:

{ x e R | x < 0}

which is a subset of R.

[spark sluice]

oh i see

[sacred sparrow]

yeah :D

[spark sluice]

it goes to -infinity

[sacred sparrow]

yop

a little tricky

[spark sluice]

Thank you very much !

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[viscid thunder]

For part a, I got upto a_2 = 2k+6 and a_3 = 4k+18 but idk what to do now

[full forum]

@viscid thunder Has your question been resolved?

[pliant vigil]

!15m

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[viral zephyr]

how to approch this question?

[livid scarab]

Each student is either in the photo or they’re not. So for each of the 6 students there are two possibilities. So how many possibilities does that make? And the don’t forget to subtract out the one case where no students are in the picture

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@viral zephyr Has your question been resolved?

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[fleet tapir]

I had this question on an exam today: The function y=25-2x has the range 13<=y<19. What is the domain of the function?

[thick hedge]

what do you think

[fleet tapir]

13<=25-2x

2x<=12
x<=6

ops

[solemn gate]

12/2 is 6 my guy

[fleet tapir]

there we go

So x is less than or equal to 6

[solemn gate]

what about y<19 part?

[fleet tapir]

25-2x<19

6< 2x
3<x

So the answer is 3<x<=6 ?

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[teal thicket]

is there an elegant way of proving this, or do you just brute force it by expansion?

[fast peak]

consider the linear independence of the columns

[teal thicket]

could you elaborate a little?

[fast peak]

what do you know about the relationship between linear independence of columns and the determinant

[teal thicket]

I'll tell you what I know and then maybe that has some relation with your statement

[torn jolt]

the first 3 columns can really only have 2 pivots

[teal thicket]

i know that i can find the matrix with cofactor expansion and if it is a 3*3 then you can find it with the triangle rule

[fast peak]

are those the only things you know about determinants?

[teal thicket]

well i have looked up some videos, which describe determinants as the factor of which an area grows from 1,1 using a given matrix

but that they can be used all over the place

[fast peak]

do you know what an invertible matrix is?

[teal thicket]

yes

[fast peak]

do you know a relationship between invertible matrices and determinants?

[teal thicket]

that does ring a bell

but i couldnt give a definition

[fast peak]

a matrix is invertible if and only if its determinant is nonzero

do you know a relationship between invertible matrices and the linear independence of its columns?

[teal thicket]

not really

so if I would like to solve this problem I need to dive deeper into the theory then?

[fast peak]

yes

[teal thicket]

so you wouldn't be able to solve this problem with the basic knowledge i have

[teal thicket]

as in this

[fast peak]

actually you could also switch rows 1<->4, 2<->5. then you have a block triangular matrix with a zero block on the diagonal

assuming you know about block triangular matrices

[teal thicket]

well

no

lol

thanks for the help

[teal thicket]

is that 2 by 2 of threes a block triangle of 3?

just looked it up

[fast peak]

no

do you know what a triangular matrix is?

[teal thicket]

its where you have a triangle in the matrix along half the diagonal

a triangle of 0

[fast peak]

you might mean the correct thing but not sure

https://i.imgur.com/WXFiwOk.png

for example this is a triangular matrix

[teal thicket]

okay

and then a block triangle is

where that r2 c2 3 is a zero?

[fast peak]

https://i.imgur.com/5flkdpa.png

[teal thicket]

i hope this isnt asking to much, but could you tell me how you can find the determinant using that?

or at least have a condition where you know that det is 0

[fast peak]

the determinant of a "normal" triangular matrix is the product of the diagonal entries

for block triangular matrices nearly the same is true

the determinant is the product of the determinants of the blocks on the diagonal

and the middle block is just a zero

so the total determinant is also zero

[teal thicket]

so the triangle rule just like a three by three matrix?

[fast peak]

if you call that triangle rule, yes

[patent forge]

if a matrix is in upper or lower triangular form and there's a 0 along the diagonal the det = 0

[teal thicket]

but isnt that p and a disturbing it?

[fast peak]

yes

but its a block triangular matrix so it still kinda works out. tho you would need to argue that thats fine

[teal thicket]

so that 0 at the centre can account for the a and p

if that is a way of putting it

[fast peak]

not really

[teal thicket]

oh

but either way if you fully calculate by cofactors and then cofactor of those you will get 0?

although it would take ages

[fast peak]

yes eventually everything would simplify down to zero

it shouldnt actually take that long

[teal thicket]

well a lot of valuable paper space at the least

well i have been occupying this chat for long enough

i suppose i will have to do some more learning

thanks for the help everyone

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[broken tangle]

If I have 2 subspaces of $\mbb{R}^4$ which I've expressed both with their span how can I find the basis of $W\cap U$

[glossy valve]

horizon2.0

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@broken tangle Has your question been resolved?

[broken tangle]

<@&286206848099549185>

[full forum]

@broken tangle Has your question been resolved?

[torn gust]

I'd set it up as you're trying to find a vector in U that fits in W, that gives you x(...)+y(...)=s(...)+t(...)+r(...) which is 4 equations and 5 variables

but since things are vector spaces, any constant times a vector in U that fits would still fit, so you can divide out by x to get 1*(...)+c(...)=s(...)+t(...)+r(...) and solve this system however you want

[broken tangle]

alright but how do i actually do it?

put the vectors into a matrix? something else?

[torn gust]

yea you put it in a matrix and maybe get a calculator to row reduce it

I'd probably just bring all the vectors over to one side and set it equal to (1 0 -3 0) the vector without a variable

[broken tangle]

im sorry but how exactly i put it in a matrix? s+2r=x, t=r+x, y=-6x this is what i get how do i continue?

[torn gust]

cool so I think you just have to look at the y=-6x part

the original premise was some combination of U that fits in W, and now you have that (1 0 -3 0) - 6*(0 1 1 0) works

that's just the basis vector of the intersection, that 1d space

[broken tangle]

i dont understand how did you find that out? and isn't it the other way around? since x is the coefficient of (1 0 -3 0)

i would appriciate if you will be able to explain in a general way or link me to a source/video explaining this procedure

[torn gust]

there's not much online for it besides https://en.wikipedia.org/wiki/Zassenhaus_algorithm

the way I stated it there shouldn't have been an x variable at all it'd just be 1, but that doesn't really matter. y=-6x means that whatever x's coefficient is, y is -6* more

[broken tangle]

😵‍💫

I don't understand a thing there lol

[torn gust]

yea that one's just an even bigger matrix

in general you're just setting up a matrix equation W=U and looking at how the variables for one of the subspaces are related to each other after you row reduce

[broken tangle]

Alright but then why don't I take into account all of the relations I've found?

[broken tangle]

Here

Why only the y and x one?

[torn gust]

they say something but it doesn't matter much for the question

it's telling you that the vector in U given in the W basis has those s t r coefficients

but we just care that it's inside the space, not how to calculate it really

[broken tangle]

But then when I get this question in a test how will I know what relation to take into account?

[torn gust]

if you set it up the same way it'd always be the bottom rows

like if it was a 2d intersection instead you'd have all 0's and if there was no intersection it wouldn't reduce

[broken tangle]

Oh I think I get it, if for example I only had 2 terms then I would have 2 vectors in the base?

[broken tangle]

And I'm still not sure I understand that, why are the vectors switched?

[torn gust]

I get it from plugging in numbers to see how it'd look

like x=1 makes the equation x=1, y=-6

like mathematically it's saying y is bigger than x right

[broken tangle]

Alright I think I understand it now

Thank you

It was very difficult😅

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[swift gyro]

Wanted a quick sanity check on whether this is true in general

[swift gyro]

where we switch x_1 and x_2 in functions f and g (but not h)

this is a generalisation of a physics problem im working on so lmk if anything needs to be clearer, from my understanding as long as x_1 and x_2 have the same limits, this would hold true?

[torn gust]

nah, the h would mess it up in general

[fathom saddle]

It's true when h is a constant though

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@swift gyro Has your question been resolved?

[swift gyro]

even if you aren't switching x_1 and x_2 in h, only in f and g?

[torn gust]

switching in h would fix it

[swift gyro]

ohhhhh i see what you mean

thanks guys ❤️

i suspect my prof forgot to put an integral on my problem sheet haha

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[dark ember]

Part 3 and 4

[dark ember]

<@&286206848099549185>

<@&286206848099549185>

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[brittle steeple]

<@&268886789983436800>

[calm nacelle]

That doesn't mean you can ping Helpers every few minutes

Please read the rules

Plus, flooding your channel with Helpers pings makes the question harder to find, so you're shooting yourself in the foot here a bit, I'll clean up the pings but please stop

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[lucid tundra]

does anyone how to solve this?

[livid scarab]

are you familiar with the golden ratio?

[lucid tundra]

i know what it is

but no i dont know much of its properties or like

special formulas ig

[livid scarab]

okay then I'll explain without really assuming anything about the golden ratio. if we suppose that the interval has length $a$, then the greater part has length $aR$ and the lesser part has length $aR^2$ (this makes all the ratios line up as described). but the whole interval is the sum of the two parts so that means $a = aR + aR^2 \implies R + R^2 = 1$. manipulating this equation should give you all the information you need

[glossy valve]

Awesam

[livid scarab]

in particular, $R = \frac{1}{\phi}$ where $\phi$ is the golden ratio. whenever you see this property of the ratio between parts being equal to the ratio between the part and the whole, the golden ratio is probably involved

[glossy valve]

Awesam

[livid scarab]

and the golden ratio has the properties that $\frac{1}{\phi} = \phi - 1$ and $\phi^2 = \phi + 1$

[glossy valve]

Awesam

[lucid tundra]

ohh

i think i c

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[wicked vine]

I have no prior knowledge involving precalc but I have to learn for a class. The word problem is: Nahee is running around the circular track with equation x^2+y^2=8100 at a speed of 4.8 meters per second. She starts at the point (0,90) and runs counterclockwise. What are Nahee's coordinates after she has been running for 27 minutes? I have tried a couple different equation but I usually end up with something like (90cos(89.6446512),90sin(89.6446512))

[wicked vine]

<@&286206848099549185>

[carmine cargo]

If you graph it out it will be easier to visualize

[wicked vine]

I have I do not know what I am looking for, I know she ends in the 4th quadrant 270.355 degrees

then the next step was reference so 360 - that = 89.6447

[carmine cargo]

make it negative because its clockwise now

[wicked vine]

make what negative?

[carmine cargo]

its 270-360

because you need to go clockwise to get to the terminal line

and when you go clockwise the degree becomes negative

[wicked vine]

okay Ill see if that works

I did not work but I have to leave thank you for trying to help

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[shrewd bluff]

Can somebody help me with the boolean algebra

[shrewd bluff]

How can I identify the ncessary format with using the NAN and NOR gates for these functions as here is a phot ogf the formula supposed to be used for the problme.

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[echo merlin]

PLEASE EXPLAIN THIS QUESTION TO ME: Determine the values of x that cause the function to be (a) zero, (b) undefined, (c) positive, and (d)
negative. SOLVE (C)

[glacial pasture]

what dont you understand

[echo merlin]

how the x value is only x=3 and not x=-5/2 as well

so part a

[sly nacelle]

Set them equal to 0

Right?

[glacial pasture]

because x cant be -5/2

or the root will have a negative argument

[echo merlin]

oh

ok

thx

[glacial pasture]

x>=3 and x not =4 is the domain of that one

[echo merlin]

ok

I know the answers I just needed help understanding

thx btw

[glacial pasture]

no worries

[echo merlin]

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[quartz ember]

I have no clue how to do this

[quartz ember]

Do i need to split the logarithm and do telescoping series??

[fathom saddle]

ye

[quartz ember]

It ended up diverging. My original solution was splitting it up and doing integral test and i got convergent which was wrong apparently. any idea why?

[fathom saddle]

Integral test should work, but it should give you diverging

[quartz ember]

okay thanks

close

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[harsh elk]

yo

[harsh elk]

anyone here in 7th grade

[carmine cargo]

No

[harsh elk]

dang

[shy edge]

🤨

[harsh elk]

does anyone know the real reason why the circle and the square have the same area

[shy edge]

Use one channel bro

[carmine cargo]

Bro what?

[harsh elk]

ye im using this one

[shy edge]

#help-11 ?

[harsh elk]

why does hte bot pin me

[fathom saddle]

Closed. You can use this one

[harsh elk]

does it do yto all of the hello messages

[shy edge]

Wdym

Square =x²
Circle =pi r²

[harsh elk]

it keeps pinning my messages to different help channels

[harsh elk]

ok thanks

So I can only talk on one channal?

[shy edge]

Use one channel to ask ur questions

[harsh elk]

oh ok

[blissful folio]

I love circles

[harsh elk]

wait so slicing the circle into pi times radius and dividing by radius squared makes it turn into a square?

bruh

the internet is so wonky

[neon fulcrum]

huh

[harsh elk]

Idk what Im saying at this point anymore

[blissful folio]

Its alright

I don't know what I'm saying half the time either

[neon fulcrum]

me too actually

but I don't think a circle ever turns into a square

[fathom saddle]

Definitely not with a compass and straightedge

[blissful folio]

radius 1/√π

1×1 square

[neon fulcrum]

I feel like you're trolling but I'm not good enough at math to really know for sure

[blissful folio]

"I must first grasp the question to grasp the answer"

• someone no history book has ever mentioned

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[harsh elk]

Ye

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[south moat]

how do I find limits?

[jolly scaffold]

which type

[south moat]

not sure, but this is an example

[neon fulcrum]

supposed to say what ahppens when x=1

[south moat]

what happens lol, im completely lost on limits

[neon fulcrum]

the picture shows it

[jolly scaffold]

the empty circles represent that the value of the function is undefined at that point

[south moat]

so DNE?

[jolly scaffold]

no

define a limit

[south moat]

idk

[jolly scaffold]

"the closest value that a function approaches without taking an indeterminate form"

basically a limit

[south moat]

kk

[jolly scaffold]

from the graph

what would be the value of the function as x approaches 0

if it is not infinity

*x approaches 1

[south moat]

i mean on the graph it seems like when x = 0 its 1, but I dont think thats the answer

[jolly scaffold]

*as x approaches 1 not 0

my mistake

[south moat]

oh kk, is it also 1?

[jolly scaffold]

yes

[south moat]

nice

[jolly scaffold]

this is the limit of the function

as x tends to 1

[south moat]

kk

also whats the diff between these 2 functions

well 3

[jolly scaffold]

the first one implies that x approaches 2 from the left side specifically

the middle one says that x tends to 2 from the right side

the third one says it in general

[south moat]

kk, are they all the same answer since its a point?

[jolly scaffold]

no

as x approaches to 2-

the function takes 1

as x approaches to 2+

the function takes 2

[south moat]

why?

oh

[jolly scaffold]

the graph

[south moat]

i see kinda

[jolly scaffold]

is split at a point

[south moat]

wait wouldnt the filled in dot be diff

[jolly scaffold]

yes

[south moat]

would it be DNE or 2?

[jolly scaffold]

the limit x tends to 2 DNE

as the limits x tends to 2- and the limit x tends to 2+ are not the same

[south moat]

ok

so what would this be then?

bc isnt it defined at 2?

bc its filled in

[jolly scaffold]

filled in means it is defined

it also has a different meaning

that the function takes the value of the filed dot

if the two are conneced by a line

so f(2) would be the value which is filled in

at x=2

[south moat]

ok

[jolly scaffold]

so f(2)=2

[south moat]

kk so when its approaching the limit from the left

its DNE bc its defined at that point

lim x--> 2

[jolly scaffold]

when it is approaching from the left it is defined

as it (theoretically) takes 1

but the whole limit (x tends to 2 not 2+ or 2-) is DNE

as it takes one from the left

[south moat]

wait what

[jolly scaffold]

and 2 from the right

[south moat]

ok, so when its approaching from the left, its 1?

[jolly scaffold]

yes

[south moat]

bc its not defined on the line approuching from the left

oh ok

and from the right is is?

it is*

[jolly scaffold]

2

[south moat]

ok

anf F(2) is 2?

[jolly scaffold]

yes

[south moat]

ok

[jolly scaffold]

take limit as x->3

what will its value be

[south moat]

2

[jolly scaffold]

why would it be 2

[south moat]

the white dot is at (3, 2)

[jolly scaffold]

i told that the circle may have a diff meaning

[south moat]

oh

[jolly scaffold]

so ya take the filled dot

as it is not defined as (3,2)

but defined as (3,1)

[south moat]

but isnt it as x approaches 3?

[jolly scaffold]

the value is determined at x=3

as f(x)=1

if it is determined the value of the limit is f(a)

at x=a (a is a value)

[south moat]

ok so does f(3) the same as x -> 3

[jolly scaffold]

if f(3) is defined and determined

determined means that that value is given or can be calculated

[south moat]

ok, but isnt it defined there at (3, 1)

[jolly scaffold]

yes

so the lim value is the determined value

[south moat]

ok

[jolly scaffold]

if it is indeterminate

then you need to take the value that f(x) approaches

[south moat]

ok

[jolly scaffold]

did you get how to find limits as a function

like y=f(x)

[south moat]

nah

but thats also one of the questions

[jolly scaffold]

these will help you understand it better

do you need help with these or can you do it

[south moat]

bro when it comes to limits im mentally handicapped

[jolly scaffold]

hahaha

so shall I help you

[south moat]

yes pls

im just tryna get this assignment in before 12

[jolly scaffold]

take no 5

substitute the value 1 in the function

[south moat]

6/4

3/2

[jolly scaffold]

that is the limit

[south moat]

so

im just plugging in shit?

[jolly scaffold]

also isnt it 6/-4

[south moat]

yes

[jolly scaffold]

Dr=x-5=1-5=-4

[south moat]

i forgot the -

[jolly scaffold]

first you plug the valut

*value

if it is a real number

then it is the req limit

but if it is an indeterminate form

like 0/0

or infinity/infinity

or infinity-infinity

then you simplify the equation

using a suitable method

[south moat]

ok

[jolly scaffold]

or use the general limits

to find the req limit

[south moat]

ok, so at x = 5 is it DNE?

[jolly scaffold]

no

ad Dr=0

[south moat]

what is Dr?

[jolly scaffold]

the value is infinity

which is defined as infinity

so the lim is infinity

Dr=denominator

[south moat]

kk

[jolly scaffold]

any more questions

[south moat]

so hold on, the answer is 0?

or infinity

[jolly scaffold]

infinity

[south moat]

bc i dont think infinity is a valid answer to my teacher

[jolly scaffold]

why

in calculus every answer is valid

[south moat]

and also

[jolly scaffold]

isnt this correct

[fossil river]

Your teacher is incorrect. Exercise are not good

There is an infinite limit definition that means a limit can equal +/- infinity

[south moat]

kk

i just care ab the grade tho

also im in pre calc

[jolly scaffold]

ask your teacher

abt infinite limits

[south moat]

i gotta turn this in by 12 lol

but ok

[jolly scaffold]

ask him/her when you dont have an assignment

or in class

for now focus on the grade

but dont be with half assed misdom

as it can bite you

pretty hard

[south moat]

kk

ok so what is this?

[jolly scaffold]

it is B

according to the definition of continuity

which you will get after limits

[south moat]

Ok

earlier response lol

[jolly scaffold]

or a definition of limit is that the limit exists if and only if Lt x->a- f(x) =Lt x->a+ f(x) =f(a)

[south moat]

kk

how so i solve these

[jolly scaffold]

factor out the highest degree variable

and cancel it

for no 15

factor out x^2

from the Nr and Dr

seperately

[south moat]

what 😭

[jolly scaffold]

take the Nr

take x^2 as common

it would be x^2((2)- (5x^-1)-(3x^-2))

[south moat]

kk

[jolly scaffold]

do the same in the Dr

what would it be

[south moat]

img ngl im not sure what u did

[jolly scaffold]

i factored out x^2

for the -5x term if you divide it with x^2

what would you get

[south moat]

idk

[jolly scaffold]

youd get -5/x

right

[south moat]

yes?

[jolly scaffold]

what abt the -3

what will you get if you divide it with x^2

[south moat]

im not sure

[jolly scaffold]

-3/x^2

[south moat]

yes?

isnt that just saying that its -3 divided by x^2

[jolly scaffold]

yes

[south moat]

i thought you were asking for smthn else lol

[jolly scaffold]

haha

now what abt 2x^2

[south moat]

ayo bro can I ask a favor

[jolly scaffold]

what

[south moat]

would it -3/2x^2

[south moat]

its like 12 mins till it has to be turned in (its 11:48) for me, can I give you like a few questions to just answer?

[jolly scaffold]

yea

[south moat]

can i start with this lol

[jolly scaffold]

15 is 2/1 or 2

16 is infinity

[south moat]

damn shit was instant lol

[jolly scaffold]

no 9 is -9/7

10 is infinity

sry 9 is -1/7

not -9/7

[south moat]

kk lol

[jolly scaffold]

i was doing some other problem

[south moat]

its cool bro 💪

[jolly scaffold]

whose answoer was -9/7

[south moat]

?

[jolly scaffold]

have any more questions

[south moat]

not rn lol, tired asf, ive been working on pre calc questions since 11am 😭

[jolly scaffold]

haaaa

ive been working on the exact same topic for 3 days

limits and continuity

but i still have like 70 more questions to do

[south moat]

damn

wish you luck bro

which calc r u in

[jolly scaffold]

derivatives

doing these for practice

[south moat]

d/dx is gonna beat the shit outta me

[jolly scaffold]

its not really hard

just time consuming

for me at least

[south moat]

Yeah we just started on those like today

Wait yesterday

[south moat]

Just do it faster

Bro I also took my SAT today

[jolly scaffold]

im not from the US

we dont have those

[south moat]

Oh kk, well it wasn’t too bad

You just gotta read it well

[jolly scaffold]

yeah

[south moat]

They also gave us desmos for the math portion

[jolly scaffold]

just remember

and substitute the right value

thats all differentiation is

[south moat]

Kk

Ima drink a lot of math equations

[jolly scaffold]

kk bye

[south moat]

Ciao

[jolly scaffold]

enjoy yourself

im still underage

[south moat]

I’m not

[jolly scaffold]

haah

[south moat]

Ayo btw which r u form

From

[jolly scaffold]

which what

[south moat]

Country

My bad for responding late lol

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@south moat Has your question been resolved?

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[wet quiver]

<@&286206848099549185>