help-33

[marsh citrus]

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[red nimbus]

(a,b) and (b,a) have to be included

[red nimbus]

If you look up the definition you have (a,a) o (a,b) -> (a,b) and (b,a) o (a,a) -> (b,a)

[elfin berry]

bacc (unhelpful)

[marsh citrus]

@red nimbus Has your question been resolved?

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[restive silo]

Hi

[restive silo]

what is the best way to get the value of y ?

I am thinking to do long division. Can that work ?

[late geode]

multiply both sides by y,
separate terms with/out y

[restive silo]

can long division work ?

[red nimbus]

dont

[late geode]

doesn't make meaningful progress

[cerulean prairie]

No it's for polynomial....

[late geode]

i mean technically you could apply it here

try and see what happens

[restive silo]

so I should have at least 2 terms in both numerator and
denominator?

[elfin berry]

who

[restive silo]

I did. it's kinda complicated but I am curious to know if it therotically correct or not

[late geode]

yes, the algebra will still be valid

it just isn't necessarily the most efficient route

[restive silo]

whenever I have such expression we sometimes use partial fraction, long division, or multiply by conjungate

that's why I am trying to have some sense of which technique I should use

[late geode]

well here it looks like the integration component is already done
getting y alone from here is just algebra

[restive silo]

I see

thanks

[marsh citrus]

@restive silo Has your question been resolved?

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[dark shard]

bored

[hidden dawn]

atp

i have no words

[rancid geode]

what

[dark shard]

hehe

[rancid geode]

why

[dark shard]

when math becomes meth

[rancid geode]

https://tenor.com/view/homelander-devastated-watch-watching-movie-gif-14922607200232575847

[knotty trellis]

2^(1/1)= 2

not +-2

:/

[rancid geode]

principal square root

[dark shard]

but sqrts have positive and negative sides

[hidden dawn]

i have no words here as well

[proud basin]

no

[knotty trellis]

oh wait

[dark shard]

nvm feck its a first root that wouldnt work

[rancid geode]

principal square root

has only positive

[dark shard]

dammit, big math beat me again

[knotty trellis]

you didnt see anything

[hidden dawn]

i didnt

yes

[elfin berry]

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

[dark shard]

how useful is the first root of something

[rancid geode]

it simplifies a lot of things

[stoic slate]

The function sqrt(x) is only defined for positive values

[dark shard]

whats with the imaginary hate

[stoic slate]

Range and domain

And 0 ofc

[dark shard]

fine

[stoic slate]

Even over complex plane

The square root of a positive number

Is only positive

[dark shard]

.close

[marsh citrus]

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[marsh citrus]

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[manic forum]

Hello I need help with this question
I know how to find the values if they gave a specific value for a but I'm not sure how to find it for all a

[tepid temple]

you want your function to be continuous

do you know what i mean?

[manic forum]

Still I only know to check for continuity at a point

[tepid temple]

that's ok

[manic forum]

F(a) defined, lim exists they equal each other

[tepid temple]

exactly!

so what you should ask yourself is, where could your function not be continuous

there are 2 points where it could not be continuous

[manic forum]

Are they maybe negative 1 and 1?

[tepid temple]

yes!

do you know why?

[manic forum]

No I just took a hint from the function

But probably they aren't equal to each other

The limit and f(a)

[tepid temple]

that's alright

[tepid temple]

that's exactly it

if m = 100 and n = -5 then the lim x -> 1- g(x) =/= lim x -> 1+ g(x)

do you understand?

[manic forum]

Yes

[tepid temple]

great

now what should you do to find a continuous function?

[manic forum]

Find the f(a) and find limit from both sides and make them equal then solve for variables

[tepid temple]

exactly!

[manic forum]

I tried this so far

[tepid temple]

it's -m + 1 you forgot a -

[manic forum]

I've got now m=-n+2 and m=n+1

[tepid temple]

perfect!

[manic forum]

M=3

N=2

[tepid temple]

nope

because this would mean 3 = -2 + 2

which is wrong

[manic forum]

Oh yeah

[tepid temple]

what you can do now is set -n + 2 = n + 1

and then solve for n

[manic forum]

n=½

[tepid temple]

yes

now you should know how to find m

[manic forum]

m=3/2

[tepid temple]

pefect

if you want you can verify if these values are correct 🙂

[manic forum]

Wow thank you so much you were so nice
I check on desmos?

[tepid temple]

np 😄 you can check with the limits 😉

[manic forum]

Ohhhh right thanks

.close

[marsh citrus]

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[manic forum]

Hello I started with solving part a and I did it this way I'm not sure if it's correct or not

[rain oriole]

It's correct

[manic forum]

Thank you

.close

[marsh citrus]

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[rain oriole]

Good approach

[marsh citrus]

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[sharp sonnet]

i need to solve this equation. i was told that i need to multiply by (z-1) and equate to (z^5 - 1) but im not really sure why

[supple marsh]

is it solve for all real roots? or does it include unreal as well?

[sharp sonnet]

in complex

so yeah unreal too

[supple marsh]

are you aware of synthetic division and the rational root theorem?

[sharp sonnet]

nope

wait synthetic divison yes but the rational root theorem no

like i can see that z^5 +1 / (z+1) is that but why

[marsh citrus]

@sharp sonnet Has your question been resolved?

[sacred idol]

in general, $x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + ... + 1)$

you can use the formula for the sum of a geometric progression to verify

[main idol]

Did you mean z-1

[sharp sonnet]

yeah my bad

[main idol]

And z^5-1

[sharp sonnet]

so thats just a thing to know?

[trail hamlet]

sum of GP :)

[sharp sonnet]

that you can rewrite that in the following way

whats GP

[trail hamlet]

Geometric progression

2, 4, 8, 16 and so on basically

[sharp sonnet]

ah like 2^n

in a sum

[trail hamlet]

read up on it ig

[sacred idol]

$S = z^4 + z^3 + z^2 + z^1 + 1$\
$zS = z^5 + z^4 + z^3 + z^2 + z$\
So, $(z-1)S = z^5 - 1$

[trail hamlet]

btw in z^5 - 1 you ignore z=1 as a solution because we divide by z-1

and we cant divide by zero

[sacred idol]

yea

[sharp sonnet]

okayokay

[sharp sonnet]

i think im being slightly slow how do you go from second step to 3rd step

[sacred idol]

subtract second and first equation

i think i skipped too fast

[sharp sonnet]

nono its okay

i'll do that rq

yeah is see it

[sacred idol]

thats kind of how sum of gp also works

from here you can solve the equation

[sharp sonnet]

okay

[sharp sonnet]

so typically when i have a sum which has a similar structure to this i'd have to use that method

or sum of gp

[sacred idol]

yeah

something like z^6 + z^4 + z^2 + 1 = 0 maybe

[sharp sonnet]

that would be more complicated to solve in that way no?

or do i have to do the thing twice

[sacred idol]

you just multiply by z^2 to the sum instead of z

basically, you multiply by the common ratio

[sharp sonnet]

ahh okok

so it would be the same as multiplying by z twice

[sacred idol]

i dont know why were multiplying by z twice but i think youre on it

[sharp sonnet]

in the example you provided you multiplied the sum by z

and thats what i did but it didn't workout so i ahve to do it again to simplify it

so it would come back to the same thing

[sacred idol]

you mean the original question or the one i just said?

[sharp sonnet]

the one you said

[sharp sonnet]

this

[sacred idol]

thats the original question

i took z^4 + z^3 + z^2 + z + 1 as S

and then just tried cancelling the middle terms by multiplying S by z and subtracting S from it

[sharp sonnet]

yeah

[sacred idol]

so what happened

you told it didnt simplify in the first step

[sharp sonnet]

yeah, then i multiplied with z again

[sacred idol]

no, like are you sure it didnt?

did you take the same steps?

[sharp sonnet]

wait we're talking about z^6 + z^4 + z^2 + 1 = 0 or my original question

[sacred idol]

thats what im asking you

you told the original question

[sharp sonnet]

oh no i was talking about z^6 + ...

[sacred idol]

if its the other one then you would multiply by z^2 yeah

because thats the common ratio

[sharp sonnet]

would it be z^2 - z or z^2 - 1 ?

[sacred idol]

z^2 - 1

common ratio - 1

[sharp sonnet]

ah okok

so then all i would need to solve is z^5 -1

[sacred idol]

yes

and remove the solution z = 1

[sharp sonnet]

do i make z = a+bi to solve?

or z = e^(xi)

and then develop to the polar form

@sacred idol

[sharp sonnet]

.close

[marsh citrus]

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[marsh citrus]

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[crisp nexus]

is this correct?

[crisp nexus]

i had to find the max product between two numbers whose sum is 100

so I took dy/dx from the first equation and applied it on the second

but im not sure if thats how it works? the answer is correct either way :0

[rocky lark]

So y+x=100 is the constraint

Basically you have to find the maximum of xy that respects the constraint

[crisp nexus]

i understand, my concern is if my method is correct

because im taking the rate of change of y from a different equation and still got the right answer

[rocky lark]

I don't understand how you did the derivative

[crisp nexus]

I had x + y = 100 and derivated in respect to x

so I got 1 + dy/dx = 0

dy/dx = -1

then the product equation

p = xy

dp/dx = y + x (dy/dx)

so I already calculated dy/dx as -1

so dp/dx = y - x

and went from there

[rocky lark]

Oh okay

So it's as if I derived x*(100-x)

That's correct

👍

[crisp nexus]

😄

thanks

.close

[marsh citrus]

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[marsh citrus]

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[steel imp]

Hello, how should I approach solving this problem?

[steel imp]

my bad, forgot about 4 above the x 🫠

although it doesnt change anything for me cuz Im as confused as I was 5 min ago

[fleet fossil]

Maybe you could post a picture of the original problem?

[steel imp]

I just have to solve this

[fleet fossil]

And you’re 100% certain that this is the problem as it is stated?

[steel imp]

yes, 100%

[fleet fossil]

<@&268886789983436800>

[peak fiber]

dealt with

[keen stump]

my thought is set each side of the (in(?))equation equal to something and solve individually

I'm still thinking of how that would specifically work

alright I've figured out how to do it

I think

we'll walk through it together lol

notice 2^(x-1) = (2^x)/2

[remote granite]

what do you need to do with it?

[steel imp]

just solve it

[remote granite]

isnt it a statement?

[steel imp]

find all x's that match the equation

[remote granite]

there's (x+3)/(9-x^4)

so 9-x^4 can't be equal to 0

$9-x^4 \neq 0$

[elfin berry]

General_Jacob

[remote granite]

$-x^4 \neq -9$

[elfin berry]

General_Jacob

[remote granite]

$x^4 \neq 9$

[elfin berry]

General_Jacob

[remote granite]

$x \neq \sqrt[4]{9}$

[elfin berry]

General_Jacob

[remote granite]

$x \neq \pm \sqrt{3}$

[elfin berry]

General_Jacob

[steel imp]

ok, you found the domain, what now?

[remote granite]

thats the solution

every x different than +- the square root of 3

hm

[steel imp]

idk man

[keen stump]

thats wacky

those decimals probably come from logarithms

[steel imp]

yeah but General_Jacob's answer is incorrect

at least thats what I think

[keen stump]

hes correct that its the domain

but I'm not sure that's what they're looking for

can I show you something that might help?

[steel imp]

yes please

[keen stump]

I'm not sure it will entirely work

this is a difficult question

[steel imp]

anything is better than nothing

[keen stump]

notice 2^(x-1)=(2^x)/(2)

simplify 3x by multiply it by 2/2

$2^{x-1}+3x=\frac{2^x}{2}+3x=\frac{2^x}{2}+\frac{2\cdot3x}{2}=\frac{2^x+6x}{2}$

[elfin berry]

CrEpasPmkinPie

[keen stump]

does that make sense so far

[steel imp]

yeah

[keen stump]

replace the right side of the equation with that new term

$\frac{x+3}{9-x^4}\le\frac{2^x+6x}{2}$

[steel imp]

hmmm, Im gonna give it a shot

[elfin berry]

CrEpasPmkinPie

[steel imp]

its probably a skill issue on my side but it leads to nothing 🫠

[remote granite]

$\log_{2}(\frac{2x+6}{9-x^4} -6x) \le \log_2 (2^x)$$

[keen stump]

$\log_{2}$ or $\log_2$ is what you're looking for

[elfin berry]

CrEpasPmkinPie

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

[remote granite]

$\log_{2}(\frac{2x+6}{9-x^4} -6x) \le x$

[elfin berry]

General_Jacob

[steel imp]

and what now 🫠

[remote granite]

can you show the original question

[steel imp]

find all x's that match the equation

[remote granite]

show it

[steel imp]

wdym

[remote granite]

show the original question

[steel imp]

??

[remote granite]

its not a question its an equation

show the question

[steel imp]

exactly, its an equation I have to solve

[remote granite]

show what they asked you to do

[steel imp]

its not in english

[remote granite]

it doesn't matter

just show it

[steel imp]

ok

[remote granite]

przez liczb....?

[steel imp]

oh, mb, "liczby"

[remote granite]

show the whole thing

[steel imp]

thats the whole thing

Im kinda scared rn 😅

[keen stump]

this is like four concepts combined into one problem

there should not be this much trouble with it 💀

[steel imp]

yeah and Im too stupid to solve it

thats why Im asking here

Ig Ill just have to live without knowing the real answer xd

[remote granite]

where did you get this problem from?

[steel imp]

from my teacher

[remote granite]

isn't it a problem from a book or something?

[steel imp]

idk, its from a work sheet assigned to me as a homework

[remote granite]

every x > sqrt(3)

[steel imp]

and some mystery numbers

[keen stump]

what is the class called?

or what grade are you in?

[steel imp]

1st year cs student

just help me without asking me personal questions

[keen stump]

it helps me understand where I should look at this from

like if this were in an algebra course differs from if it were a calculus course

[steel imp]

oh, Im sorry, for some reason I thought youre jacob xd

[steel imp]

understandable

.close

[marsh citrus]

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[simple leaf]

I'm in basic calc 1, and we are currently covering second derivatives. How do I prove this?

[simple leaf]

I've tried taking the derivative of the definition of the derivative, but it doesn't seem to work

<@&286206848099549185>

<@&286206848099549185>

[remote granite]

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

[elfin berry]

General_Jacob

[remote granite]

$f''(x) = \lim_{g\rightarrow 0} \frac{f'(x+g)-f'(x)}{g}$

this might help

[elfin berry]

General_Jacob

[remote granite]

$f''(x) = \lim_{g\rightarrow 0} \frac{\lim_{h \rightarrow 0} \frac{f(x+g+h)-f(x+g)}{h}-\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}}{g}$

[elfin berry]

General_Jacob

[simple leaf]

why is it g and not h?

[remote granite]

why can't it be g?

[remote granite]

g and h are approaching 0 so you can use just one letter

$f''(x) = \lim_{h\rightarrow 0} \frac{ \frac{f(x+h+h)-f(x+h)}{h}- \frac{f(x+h)-f(x)}{h}}{h}$

[elfin berry]

General_Jacob

[remote granite]

$f''(x) = \lim_{h\rightarrow 0} \frac{f(x+h+h)-f(x+h)}{h^2}- \frac{f(x+h)-f(x)}{h^2}$

[elfin berry]

General_Jacob

[remote granite]

hmmm

it doesn't seem right

[simple leaf]

yeah

I'm stuck on that aspect as well

[remote granite]

$f(x+h+h)-f(x+h) = f(x+h)-f(x)$

[elfin berry]

General_Jacob

[remote granite]

do you think it would be correct to do this?

i think i know what you can do

[simple leaf]

really?

[remote granite]

f(x+g+h)-f(x+g)

you don't have to add additional offset g when there is already h

[simple leaf]

wait why?

[remote granite]

so you can just write f(x+h)-f(x)

[simple leaf]

wait even so

how do you get -2f(x)

[remote granite]

hmmm

[simple leaf]

wait can you explain the above equation

im not exactly following it

[remote granite]

this is wrong i think

forget it

[simple leaf]

um okay

how should i approach this then

[remote granite]

did you learn about taylor expansion?

[marsh citrus]

@simple leaf Has your question been resolved?

[simple leaf]

no i havent

[marsh citrus]

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[desert wharf]

I've tried but can only proof it is true

[desert wharf]

<@&286206848099549185>

[marsh citrus]

@desert wharf Has your question been resolved?

[marsh citrus]

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[ashen mulch]

I'm doing linear algebra. I'm working on the chapter "coordinates", subsection "coordinate transformations". I have no idea what this means.

[ashen mulch]

It says: Given ( f : V \to W ) with ( n = \dim V ) and ( m = \dim W ). Then there exists a basis ( \mathcal{B} ) of ( V ) and a basis ( \mathcal{C} ) of ( W ) such that
[
A_{f, \mathcal{B}, \mathcal{C}} =
\begin{pmatrix}
I_k & 0_{k, n - k} \
0_{m - k, k} & 0_{m - k, n - k}
\end{pmatrix}
]
for a certain ( k \leq n, m ).

[elfin berry]

Nico

[ashen mulch]

What does this mean 😭

<@&286206848099549185>

[marsh citrus]

@ashen mulch Has your question been resolved?

[marsh citrus]

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[split vigil]

anyone care to explain?

[marsh citrus]

@split vigil Has your question been resolved?

[full flume]

any part in particular you're stuck on?

[split vigil]

I mean i understand a if that helps 😭

[vale panther]

have you solved a?

[split vigil]

365^n right

[full flume]

yep

[split vigil]

Ye

I thought b was 365!

[full flume]

so that's like choosing dates from 365 with replacement

and b is going to involve factorial yea, but we'll work up to it

for b we want to choose n dates. I like to think of it like picking them in order

once one is chosen, we repeat but can only use the dates that are left

[split vigil]

Got that part

[full flume]

cool

so for one person we have 365 choices

then the second, 364

and so on...

get that in terms of n

[split vigil]

Yeah

[vale panther]

have you heard of permutation?

@split vigil

basically that

[split vigil]

n then n-1 n-2 until ......... 1

[vale panther]

not quite

[split vigil]

I guess but im a newbie in this

[vale panther]

365! is correct if n = 365

[split vigil]

Oh

[vale panther]

but in general if n is not 365 how would you construct an equation

[full flume]

good link here for reference https://simple.wikipedia.org/wiki/Permutation

[split vigil]

Is the permutation of 4 people sitting on 4 seets

4x3x2x1

[full flume]

yes

[vale panther]

what about 4 people sitting on 2 seats

[split vigil]

4x3

[vale panther]

👍

what about 365 dates for every n students?

[split vigil]

365 x 364 .......... x n ?

[vale panther]

how would you express this in general

[vale panther]

slight mistake

[vale panther]

did you do 4 x 3 x 2 here?

[split vigil]

Not really

The first seet would have a possibility that 4 different ppl can sit on it

The second seet would only have 3

[vale panther]

right

[split vigil]

Because 1 of those 4 are already seated

[vale panther]

the first student has a possibility of 365 dates

[split vigil]

Yeah

[vale panther]

second student 364

last student = ?

[split vigil]

You spoiled it a bit didnt u

Would be n

[vale panther]

no?

[split vigil]

365 - n

[vale panther]

yes

so the n-th student as 365-n dates to choose from

[split vigil]

365 x 364 ......... x 365-n

[vale panther]

exactly

you got it

[split vigil]

Oh thats it

Hmm

[vale panther]

the formula would be 365! / (365 - n)!

[split vigil]

Uhm

[vale panther]

which is the same as 365 x 364 x ... x 365 - n

[split vigil]

Yeah okay

True

[vale panther]

yes?

[split vigil]

Okay that does clear things up

How would c work?

[full flume]

quick correction! this would be 365 * 364 * ... * (365 - n + 1)

[vale panther]

right 😭

my bad

its safer to use the permutation formula

sometimes u slipped if you tried to use intuition and forget the +1

😭😭

[full flume]

frfr i always have to triple check and plug in dummy values whenever i divide factorials

[vale panther]

oops

hmm how do I teach this without giving too much

[split vigil]

Is it too simple

Like explaining how + and - work

[vale panther]

I gave away the main point of the question lol

😭😭

hope u didnt see it

[split vigil]

I get P(no matching) = 1 - P(matching) if u explain me how i can fight P(matching)

[vale panther]

oh thats just

[split vigil]

Find*

[vale panther]

if x is a date

from 1-365

and y is a date from 1-365

what is the probability that x = y

[split vigil]

No idea

[vale panther]

@split vigil can you tell me what the definition of probability again

[split vigil]

Probably 1 in 365 or sum

[split vigil]

Chance of something happening

Idkk

[vale panther]

formal definition

P(A) = |A| / |Omega|

i forgot how to use latex in this server

[split vigil]

First time seeing this

[vale panther]

are you in uni or highschool

[split vigil]

Uni

[vale panther]

it should be the first thing you see in textbooks

[split vigil]

Tf is omega

[vale panther]

omg its a whole thing but

omega is the set of all possible outcomes

[split vigil]

I have introduction to probability theory not the real deal yet

[vale panther]

for ex if we toss a coin

omega is {heads, tails}

[split vigil]

Ohh

Thats S

Sample space S

For me

[vale panther]

diff notation 👍

anyway

[split vigil]

Ok yeah

[vale panther]

probability is cardinality of event A over cardinality of omega

so can you tell me what is the probability that a date is 7 from 365 possoble dates

[split vigil]

7/365

Omega is 365

Event A is 7 of them

[vale panther]

@full flume help I suck at teaching

[vale panther]

no I mean that the date is 7

[split vigil]

Oh if the date is 7

[vale panther]

like 7th of the year

[split vigil]

7 januari?

[vale panther]

yea

[split vigil]

That would be 1 in 365

[vale panther]

👍👍

okay so now what if you throw two 3 sided dice

what is omega

size of omega

[split vigil]

Ill just five the dices numbers 1 2 3

Give*

[split vigil]

9?

[vale panther]

yes

what is the size of the event that the two dices display the same number

[split vigil]

Depends id you look at the total sum of the values of the dice or just yeah

[split vigil]

Out of the 9, only 3 of them would have this

So the probability is 3/9 of two dices having same number

[vale panther]

now apply it to the previous question

[split vigil]

P matching

[vale panther]

this one

[split vigil]

The chance of it matching

If x is a date, there is a x/365 chance of that event happening

[full flume]

hi i missed the ping sorry, and the dice arc 😵‍💫

[split vigil]

So 1 - x/365

Would be P(not matching)

Or what

[full flume]

wait we're doing part c ?

[vale panther]

yes

[split vigil]

Ofc

[vale panther]

I'm very bad at teaching 😭😭

[vale panther]

not quite right

[full flume]

we have the number of total possible outcomes. and we have the number of possible outcomes for some restriction. this was part a and b.

what's the probability of that restricted outcome (matching bday) ?

[split vigil]

Idk man

Its like almost 4am for me too

[full flume]

haha no worries

[vale panther]

[full flume]

so sample space size is 365^n

[split vigil]

Oh

Divide answer b by a

[full flume]

yiss lmao

sleep deprivation is no joke

[split vigil]

But i dont understand

How is b gonna determine the chance of it not being matching

[full flume]

i think don't get caught up in the "not"

there's some property which we know the number of ways it can happen

365! / (365 - n)!

so the chance of that outcome is just that divided by the total number of ways

[split vigil]

Ahh

[full flume]

lot's of negation flying around haha

[split vigil]

😂

Thank you guys

Ill be asleep within 30 secs of me closing my phone

[full flume]

bonus fun this is actually a famous example haha
https://en.wikipedia.org/wiki/Birthday_problem

sleeeeeep!!! 💤

[split vigil]

Gn

[marsh citrus]

@split vigil Has your question been resolved?

[marsh citrus]

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[raw quest]

I don't understand how to do it

tue first bit is a gp right and i tried to put it in gp formula but it didn't work

[marsh citrus]

@raw quest Has your question been resolved?

[main idol]

Show what you tried

[raw quest]

Okok

[marsh citrus]

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[raw quest]

.reopen

[marsh citrus]

✅

[raw quest]

<@&286206848099549185>

[pure jacinth]

I think your sum formula is off by 1

and there's a second error in your exponent

remember that z^2 is the common ratio, so you're not summing 2n-2 terms

after that, it's just a matter of common factoring to make your expression match the one given

[marsh citrus]

@raw quest Has your question been resolved?

[raw quest]

ohhhh

ur right okok

thsnks

[marsh citrus]

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[honest creek]

hi

[honest creek]

how does the area of an equilateral with side length 8 compare to the area of a right triangle with perimeter 25

i’m surprisingly having a bit of trouble here

[cobalt sedge]

the area of a right triangle with perimeter 25 is not unique since you can build infinite such triangles. Maybe you wanna compare the maximum possible area of such a triangle?

[honest creek]

sure

[cobalt sedge]

Max area would be from an isosceles (45-45-90) triangle

[honest creek]

is there no other way to do it?

other than knowing that factoid

[cobalt sedge]

you can write a general expression for sides and calculate the area in such way, and then maximise the resulting quadratic expression

its your choice

[honest creek]

i mean the hypotenuse would have side length 12 at its maximum

and then the other two sides have to sum to 13

[cobalt sedge]

12.5, 12.5, 0 is the triangle with 0-90-90 angles

So, with an area of 0, it would be the smallest area

you can tweak the angles a bit to get a hypotenuse slightly smaller than 12.5 (and still larger than 12)

but that area would be a very small number nonetheless

[crystal lintel]

[honest creek]

hi

[honest creek]

no kind of bounding works?

i tried A = rs as well

i treated r as the circumradius

[cobalt sedge]

wdym bounding?

[honest creek]

but that flopped too

[cobalt sedge]

r in that formula is inradius, not circumradius

[honest creek]

area of right triangle with bla bla < some value

[honest creek]

yes i’m aware

but i don’t know the inradius right

but i know the circumradius is at most 6

and circumradius is less than inradius

so i bounded above

[cobalt sedge]

so you use some unrelated quantity? wtf kinda logic is that

[honest creek]

it’s still too big to compare

[honest creek]

no i bounded it 😭

[cobalt sedge]

no?

[honest creek]

area of right triangle < (6) (12.5)

why is that not valid? the argument uses the fact that 6 is the maximum possible value of inradius (it’s probably like 3)

[honest creek]

more than*

[honest creek]

anyway this failed as well

[cobalt sedge]

circumradius is always larger than inradius. Thats by definition

[honest creek]

[honest creek]

yes

[honest creek]

i use that here

that’s what i meant by bounding above

[proud basin]

is this trying to find the max area of a right triangle with perimeter 25?

[honest creek]

semiperimetee is 12.5

[honest creek]

uhhh yep

[proud basin]

you can manipulate algebraically and use am-gm

[cobalt sedge]

There is a lot of issues with this bounding. The radii you use are dependent on the exact dimensions of the triangle. You want upper bound for area, use the area of isosceles right triangle

[honest creek]

hmm but where did you get the isosceles thing

[honest creek]

u can solve it within 1 minute? it’s meant to be solved by heuristics

no math no problem

just get the right answer which is equilateral triangle has the greater area

[honest creek]

okay so the area of a square is the maximum for a fixed perimeter right?

[cobalt sedge]

from here

[cobalt sedge]

yes.

[honest creek]

yeah where’d you know that from

what about a circle?

okay where do i find more fun facts like this 😭😭

[cobalt sedge]

you can extend this. Half the square is isosceles right triangle

[honest creek]

yes

because the diagonal is an angle bisector

[proud basin]

honestly if i got this i would guess its either a 45 45 90 or a 30 60 90 and try to compute them (if i cant then just pray) :p

[honest creek]

30 60 90 i hope is not special right?

well other than nice side length ratio

ugh okay this question is weird

[proud basin]

:shrug:

[cobalt sedge]

since square has max area, 0.5*area of square is the max half area possible of all rectangle, and that means iso right tria has largest area amongst all right trias

[honest creek]

yes i understand that now

do you know any other

things like this

which u think might be important for me

or well uh

what about a circle

given a fixed perimeter, does it achieve the maximum area?

[cobalt sedge]

that the area of a circle is max for all polygons with same perimeter?

yea, thats true

[honest creek]

yeah i guess this is true

[honest creek]

oh

[honest creek]

a square is next?

[cobalt sedge]

no?

idk?

maybe

[honest creek]

[cobalt sedge]

you can calculate it yourself

[honest creek]

what am i calculating though

idk how to prove it

and i googled it

this whole thing is pretty advanced

i guess

maybe i’ll just memorize this fun fact

it has to do with isoperimetric inequality

[cobalt sedge]

wdym advanced? its not very advanced tbh. A bit of trig and a bit of quadratic equations and you can calculate this yourself

for any polygon, the regular polygon has larger area than any other polygon with same perimeter

And you can compare areas of regular n-gons for different n using trig

[honest creek]

hmm okay sure

maybe i’ll let a computer do it

thank you

.close

[marsh citrus]

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[marsh citrus]

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[supple marsh]

How

[supple marsh]

I expanded the expression

(Multiplying brackets

How to solve next?

I found the terms with x^0 and x^-3 now I need to find a term with x^-1 to cancel out 2x

But there is no such term, I checked using the general term

[proud basin]

personally i would think like this
the "pair" with 1 on the left bracket would be a constant on the right
the pair for 2x is something/x on the right
the pair for -3x^3 is something/x^3 on teh right

you can find the ammount of constants on the right, the ammount of something/x on the right, and the something/x^3 on the right

[supple marsh]

Exactly what i did

But there is no term with 1/x...

[proud basin]

ok

hm

if it doesent exist then its 0

[supple marsh]

What

[proud basin]

well if there is no term with x^-1 then the 2x has no pairs

[supple marsh]

See the formula for calculating a general term is Tr+1=nCr a^n-r b^r

Where r and n are natural

If you sub and but -1 in the coefficients of x, then you get r which is not natural

[supple marsh]

Yes, but how is it 0

Ohh

I got it

You mean there us no constant term in that

So it should be ignored

Alright thanks

.close

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[lyric cobalt]

Hi guys I have a question about derivatives. Why do we have to calculate the derivative of a point using its definition : lim (y-y')/x-x' and not just use the tangent to that point to calculate that derivative ?

[fervent rampart]

how would you find the tangent line without taking the derivative first?

certainly if you had the equation of the tangent line that would tell you the derivative, but we don't usually know it ahead of time

[unreal oxide]

that definition is kind of equivalent to the tangent

what difference do u see between that and the tangent?

[lyric cobalt]

oh nvm I was thinking we can just draw a straight line and use the tangent but that won't work before studying the function. Thanks

[unreal oxide]

and "drawing" is not very accurate

[lyric cobalt]

yeah I understand. Ok so I have another question about limits. What is the actual use of limits ? I mean why do we need an approximation of a value while searching for the exact value ? like if lim f(x) = 2 but x isn't defined in that point then what is the point of studying f(x) as x gets closer to an x but then we can't use it because its not even defined there ? Like in real life application how could limits be useful or not cause any problems ?

I think they can be useful in things like continuity or getting the approximation of a real number using numerical sequences or something but for something that we can't get its exact value ?

[fervent rampart]

well for example when we are studying derivatives, the function [ \frac{f(x+h) - f(x)}{h} ] is not defined at $h = 0$, because we end up with $0/0$. however when $h$ is close to 0, we get the slope of the secant line. The limit as $h\to0$ allows us to find the slope of the tangent line

[elfin berry]

cloud

[lyric cobalt]

but it would be just an approximation of the tangent not the exact value ?

can it be useful in real life applications of math ?

[fervent rampart]

no, the value of the difference quotient is approximate for any finite value of h. but in the limit it becomes precise

limits are good at making approximations into precise measurements

[lyric cobalt]

but if that exact value isn't defined then what use do we have from getting its precise measurement ?

[fervent rampart]

well we saw that the slope of the tangent line is not defined if we just use the formula above. but if we take the limit we do get a precise measurement of its slope

https://www.youtube.com/watch?v=oc0M1o8tuPo

[unreal oxide]

maybe..suppose theres a path u r walking along. at just one specific point u know there's some infinitely deep hole or something. u dont like to step on that point but ud like to know whats happening as u approach that point..ish

then maybe u can decide if u can actually continue walking along the path if u somehow are able to avoid that one point

maybe if the path seems to be something like a "natural continuation" of what uve seen so far, u can do that.
maybe if it looks like a sudden step, u cant, coz ull topple over if u try to avoid it

im trying to think of a practical use lol

[lyric cobalt]

ok thanks Ill check out this video

[lyric cobalt]

that was actually a really good example lol. so as we get closer to that point we want to know what would be happening but since we can't fall in that hole(isn't defined) then we can never reach it but we just study how it can affect us as we get closer to it. But then again if we can never fall then what is the point of studying how would we fall, or when would we fall, or what would happen if we fall ... ?

and if we would just LIKE to know that info, then how can we use it ? If that info is originated from somthing that doesn't exist and will never exist

[unreal oxide]

maybe u r driving a car and u have a path like this

with a hole in the way

say u can still drive over the hole without falling into it

but if the path looked ilke this

u cant. ud topple over

[unreal oxide]

here the limit exists

[unreal oxide]

here it doesnt

[unreal oxide]

so this path isnt really any different from one that has no hole in it at that point (like from the pov of the car). it can just drive over it

if u wanted to analyze this path for some reason, maybe to find the speed of the car or whatever, u dont want to go into the pain of working with a function defined in a strange way like that, with a hole

instead u define a new continuous function, without a hole, where the value at that hole point is equal to the limit there. so u will need to evaluate that limit and find the value

and now its easier to work with that function

[marsh citrus]

@lyric cobalt Has your question been resolved?

[marsh citrus]

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[lyric cobalt]

ok thank you this made things more clear for me

[lyric cobalt]

can anyone help with the first question ? in the second limit there is a mistake, x tends to minus infinity not +infinity

F continous in IR, lim(x->+ ♾)f(x) = l and lim(x->- ♾) f(x) = l’

Show that if ll’<0 then there exist a,b in IR in which f(a)*f(b) < 0

[unreal oxide]

So it's a typo in the pic? It must be negative infinity right

[lyric cobalt]

yes

yes for the second limit

[unreal oxide]

I'm thinkig something like the intermediate value theorem

[lyric cobalt]

btw I solved it using the definition of a limit I just want to know if there is another way to solve it

[lyric cobalt]

yeah this question is to prove the intermediate value theorem

so you use it in the second question

[marsh citrus]

@lyric cobalt Has your question been resolved?

[marsh citrus]

@lyric cobalt Has your question been resolved?

[marsh citrus]

@lyric cobalt Has your question been resolved?

[marsh citrus]

@lyric cobalt Has your question been resolved?

[lyric cobalt]

<@&286206848099549185> help pls

[royal sable]

That's probably the best way, to be honest

Definition of the limit and ivt

Takes like 2 lines

[lyric cobalt]

we can't use ivt, in this question we must show that ivt exists so we can't use it. but ok I guess the definition is the only way. Thanks

.close

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[round marlin]

How do I convert my answer to the one in the textbook? trying to find the zeros of 2x^2 + 4x - 3

[round marlin]

Threw them into a calculator and they’re the same value but I would like mine to look the same as the one in the book cause my professor isn’t chill like that

[near fossil]

does this make sense?

i think it is just to make it look better as there is no fraction under the radical

[round marlin]

That makes sense

Thank you sm

[near fossil]

no problem

[stoic slate]

Multiplying by 1 =2/2 the sqr(5/2) also works

[round marlin]

interesting

ty

[lyric cobalt]

Btw I think it should be t = … “or “ t= …
It can’t be “and” because a variable can’t have two different values at the same time. When we say the solution is x=.. or x=.. then x can be one of them to satisfy the equation. But if we say “and” its like saying x must be equal to two different values at the same time to satisfy the equation which isn’t possible

[round marlin]

that's the textbook answer I didn't write that out

[lyric cobalt]

Nothing related to the question but just wanted to point it out so you don’t get used to writing “and” 👍

[near fossil]

what class is this?

[round marlin]

I usually write "or" cause it makes more sense in the context of the graph

[near fossil]

algebra 2?

[round marlin]

precalc/trig

[near fossil]

oh

[round marlin]

my professor won't let us use the quadratic formula for some reason

[lyric cobalt]

Yeah ik but just wanted to tell you in case you didn’t know

[near fossil]

i mean that’s fair not every equation is a quadratic

[lyric cobalt]

Bro how are you rading this fast you reply in a second

[round marlin]

true but its frustrating that it says to do so in the textbook and he'll mark u down for doing it

[near fossil]

that is annoying

[round marlin]

anyway I got homework to do thank you for the help

appreciate it 🙏

[lyric cobalt]

I think there are other ways to solve it other than the quadratic formula

Like using some techniques but just in certain cases

[marsh citrus]

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[marsh citrus]

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[knotty plover]

Hi

[knotty plover]

I need help with This calculus question

[marsh citrus]

@knotty plover Has your question been resolved?

[stoic escarp]

it gives you the whole perimeter is 8m

[marsh citrus]

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[knotty plover]

.reopen

[marsh citrus]

✅

[knotty plover]

@stoic escarp did u close the ticket?

[stoic escarp]

no it autoclosed because of time

[knotty plover]

Oh so can u help?

[stoic escarp]

it gives u the whole perimeter is 8m

imagine the shapes separately

[knotty plover]

I know that obviously

[stoic escarp]

the radius of a circle is from the centre to the edge

so the flat bit of each semi circle would be 2r

you can then see how two flat bits of the circle add up to make the rectangles length

[knotty plover]

So the length is 4r

?

[stoic escarp]

ya

[knotty plover]

And what is the height

[stoic escarp]

try figure it out

i gotta go sleep anyway

[knotty plover]

How?

Bruh do you even know?

[marsh citrus]

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[lean osprey]

can someone help with this one please

kind of confused

<@&286206848099549185>

[marsh citrus]

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[lean osprey]

<@&286206848099549185> anyone there

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[mystic ledge]

i belive the the theta is changing from pi/2 to 5pi/6 in the second quadrant and r from 0 to 4

[lean osprey]

hmm

im not sure

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[marsh citrus]

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

[dry musk]

1

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[cinder garnet]

.close

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[eternal flame]

(X_n, d_n) is a metric space
for all n => 1

[spark otter]

what's your question about this?

[eternal flame]

Is the question just asking to show its a metric space?

Like (X,d)

[quaint elm]

i think so yes

[eternal flame]

Okay so im proving the 3 properties but having diffuculty to show the trianglw inequality one

[marsh citrus]

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[marsh citrus]

@eternal flame Has your question been resolved?

[eternal flame]

Is this proof valid?

[spark otter]

I have no clue what happened when you do the red and green star thing to cross reference

[eternal flame]

As in what its referncing or the steps used to get there?

[spark otter]

idk but

why is this true tho?

[eternal flame]

for red i replaced dn(xn,zn) with dn(xn,yn) + dn(yn,zn)

since dn is a metric

[spark otter]

but

[eternal flame]

traingle inequality holds

[spark otter]

listen

[eternal flame]

ok

[spark otter]

while the numerator is indeed replaced by something bigger

the denominator is ALSO replaced by something bigger

[eternal flame]

Okay im with you

[spark otter]

how do we know

that the denominator is not like so much bigger now compared to the previous one

that it's "stronger" than what happens at the numerator

for example

say 0 < x < y

[eternal flame]

yes

[spark otter]

we also have x^2 < y^2

[eternal flame]

yeh

[spark otter]

does that mean x/x^2 < y/y^2?

[eternal flame]

going to be honest not sure

[spark otter]

yeah, in fact it's false

[eternal flame]

why

[spark otter]

x/x^2 is in fact always > y/y^2

x/x^2 = 1/x

y/y^2 = 1/y

[eternal flame]

ah yes

sorry i get that

[spark otter]

inverse function decreasing on (0,+inf)

so

[eternal flame]

okay so im done that wrong

[spark otter]

not apparently true

don't get me wrong, you need to end up to be able to say that

you're just gonna have to prove that dn/(1+dn) is a metric

hint: study differentiability and variations of f(t) = t/(1+t)

[eternal flame]

So I still need to arrive at the same result which would be valid if 1+do is a metric

Dn not do

How would you show that it is like briefly with proving all the properties again?

[marsh citrus]

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[urban lodge]

I can’t really figure out where I went wrong lol

[urban lodge]

The answer is 60 and 240

Yeah nvm

[tough igloo]

u wrote alpha = -30 degrees, but u wrote ur equ as sin(2x+30), should be sin(2x-30)

[urban lodge]

The tangent part I got an additional negative in

[tough igloo]

and it should fix up everything

[urban lodge]

lol

[urban lodge]

Yeah I was fixing up another question when I also realised I have been putting an additional negative for like the past 3 questions if the second term ends in - lol

I am so cooookkked

[tough igloo]

lol its okay

[tough igloo]

u got thissss!!

[vivid cipher]

Take a break then!

[tough igloo]

goodluck!

🫡

[vivid cipher]

Have a great day!

[urban lodge]

I think I will. Been grinding for like 5 hours lol

Anyways thanks lol

.flose

[urban lodge]

You too!

.close

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[young burrow]

evening may i know how 2 draw the ambigious im quite bad at it 😔

[marsh citrus]

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[still temple]

youre gonna need to translate this

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