help-38

[solid kiln]

lILi

[quick hound]

ye that

[dreamy sleet]

,tex $\ln(4^y)=y\ln(4)$

[solid kiln]

lILi

[dreamy sleet]

for this, they exponentiate both sides with a base of 4 to get rid of the logarithm, take the natural log of both sides to bring out the y again, divide both sides by the natural log of 4 to isolate the y, and differentiate

[delicate lance]

ohh

[dreamy sleet]

If you give me a moment, I can type it up more clearly

[delicate lance]

I don’t get this part tho?

[dreamy sleet]

dividing by ln4 is the same as multiplying by 1/(ln4)

[quick hound]

ln 4 is just a value

[dreamy sleet]

,tex
\begin{enumerate}
\item $y=\log_4(6x^5-8x^2+11)$
\item exponentiate both sides: $4^y=6x^5-8x^2+11$
\item take natural log of both sides: $y\cdot\ln(4)=\ln(6x^5-8x^2+11)$
\item Isolate y: $y=\frac{1}{\ln(4)}(\ln(6x^5-8x^2+11))$
\end{enumerate}

[solid kiln]

lILi

[delicate lance]

no way u typed all this out

wtff

[quick hound]

holy speed wtf

[delicate lance]

holy moly

bros a human calculator

[quasi saffron]

bro is a certified bot

[dreamy sleet]

I'm taking an 8-week calculus 2 course so I taught myself how to take notes and do my homework with LaTeX to save myself the pain of writing it all out by hand. From all the note taking, i've become quite proficient at writing out LaTeX code

[quick hound]

your proff accepts latex?

[delicate lance]

so moving ln4 to the other side means converting it to 1/ln4

[dreamy sleet]

Yeah he's cool like that

[delicate lance]

oh

[quick hound]

ln 4 is just a value

just imagine its 5

how do you move it to the other side

[dreamy sleet]

we're diving both sides by ln4

yeah

[delicate lance]

subtract it from other sidde

wait

[quasi saffron]

yeah basically u want to move it but in math ways you need to eliminate the other side by adding the one side

[delicate lance]

i have to divide the whole equation by it

[quick hound]

yes

[delicate lance]

me when the whole server helps me a day before my calc exam

[delicate lance]

Are u going to be alive tmrw to help me

[quick hound]

i guess

what time is tomorrow

[delicate lance]

buddies now

[quick hound]

cool

[delicate lance]

Idk like 5 pm Idk

[dreamy sleet]

@quick hound If you can help her with the rest of this, I can start typing up the basics of latex to send to you, ping me if you need anything written out

[delicate lance]

wow ur so helpful

become a teacher

i feel so supported

[dreamy sleet]

I want to be a rocket propulsion engineer but I might end up teaching one day

[quick hound]

sure

thats cool

[delicate lance]

that’s insane

i want to be a game developer

rocket stuff is insane

[trim jolt]

@delicate lance Has your question been resolved?

[dreamy sleet]

Alright, i'm done typing that all out

[dreamy sleet]

Is there anything else you need help with?

[delicate lance]

Yes please

[delicate lance]

Can you

Do you know optimization

[dreamy sleet]

Probably enough to help you yeah

[delicate lance]

Okay

[dreamy sleet]

Typing out the prompt:

A rectangular poster has an inner rectangle printing area and margins of 5cm on the top and bottom, as well as margins of 3cm on the left and right sides. If the poster has an overall area of 9375 cm^2 (including the inner rectangular printing area and the margins all around it), then what is the maximum possible area of the inner rectangular printing area?

• You do not need to include a domain analysis for this question. Round your answer to one decimal place if necessary.

So, we can define the total area and the printable area as such:

,tex
$\begin{cases}
A_{\mathrm{tot}}=x\cdot y=9375\
A_{\mathrm{p}}=(x-6)(y-10)
\end{cases}$

[solid kiln]

lILi

[delicate lance]

my question is, why do they do (w-6)(l-10) instead of (w + 6)(l + 10)

[dreamy sleet]

they're solving for the printable area

which is the total width and length minus the margins

[delicate lance]

ohhhh

[dreamy sleet]

we can define y in terms of x

[delicate lance]

how did they solve for domain

[delicate lance]

oh

[dreamy sleet]

they aren't

remember, they said they don't need to analyze the domain

[delicate lance]

oh ok

[dreamy sleet]

$y=\frac{9375}{x}$

[solid kiln]

lILi

[dreamy sleet]

Let's expand Ap

,tex $A_p=xy-10x-6y+60$

[solid kiln]

lILi

[dreamy sleet]

now we replace y with our definition of y

,tex $A_p=x\cdot\frac{9375}{x}-10x-6\cdot\frac{9375}{x}+60$

[solid kiln]

lILi

[dreamy sleet]

,tex $=9375-10x-\frac{56250}{x}+60$

[solid kiln]

lILi

[dreamy sleet]

,tex $=9435-10x-56250x^{-1}$

[delicate lance]

yes yes

[dreamy sleet]

If we assume that there's only one extrema (the maximum), we can solve for where the derivative equals 0

but first, we need to find the derivative

so go ahead and differentiate that equation

[delicate lance]

i understand everything now

i don’t need to do

i’ll ask another question

[solid kiln]

lILi

[dreamy sleet]

just had to fix that rq

[dreamy sleet]

go ahead

[trim jolt]

Channel closed

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[delicate lance]

wait so

they’re minimizing surface area

[trim jolt]

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[delicate lance]

so any extra info like the volume is used to solve for one variable

[dreamy sleet]

yes

[delicate lance]

ok i get it

[dreamy sleet]

yeah you're solving for the radius that will minimize surface area

[delicate lance]

how an this question

i’m screwed for the exam

there’s no way

how is this just one unit

[quasi saffron]

bro u just have to be confident

[delicate lance]

alt

alright

i’m gonna get an 80

[dreamy sleet]

I need to refresh myself on inscribed shapes in elipses so you'll need to get someone else for this

[delicate lance]

its ok

how an

question 3

tell me why

do i need a new pathway

[dreamy sleet]

Ok this is something I can do

[delicate lance]

fa becomes fe + eb + ba

[dreamy sleet]

give me a moment to draw something out

[delicate lance]

dc becomes de + eb + bc

dc + fa |

fe + ed cancel eachother ou

becoming 0

eb + eb becomes 2eb

[dreamy sleet]

just a moment

[delicate lance]

okay

[quasi saffron]

is it o its a rotation?

[quasi saffron]

this once be my fave but now i kinda rusty with it

[delicate lance]

how the hell

this was my least favorite

how was it EVRR your favoritr

[quasi saffron]

limit and derivatives

haha

[delicate lance]

trig limits dude

[quasi saffron]

which one is your confusion

[delicate lance]

all

what’s the goal

when solving

[quasi saffron]

ah

[delicate lance]

ever since that spider flew on my keyboard i keep feeling little hairs everywhere

[quasi saffron]

u need to move and get something clean

wait lemme find it

so limit is basically try to find the closest to the x

in that case is x to 0

[delicate lance]

oh ok

[quasi saffron]

wait the IILi god is typing

[dreamy sleet]

FA is a vector defined as the sum of vectors FE, EB, and BA
DC is a vector defines as the sum of vectors DE, EB, and BC
DE and FE are opposing vectors, so are BC and BA, so they can be defined with:
DE=-FE, BC=-BA
FA+DC=FE+EB+BA-FE+EB-BA=EB+EB=2EB

[delicate lance]

ohhhhhhh

it makes more sense

i just realized

de eb and bc is just dc but in a long way

its a new path

[dreamy sleet]

to elaborate on what I mean by being the sum of vectors. FA starts at point F, moves right along FE, moves up along EB, and then moves left again along BA

DC does the same but along DE, EB, and BC; and in the opposite direction

[delicate lance]

ohh

do i have to draw arrows on the diagram

for marks

[dreamy sleet]

so, look at the equations they wrote out

they defined vector FA as FE+EB+BA

and DC ad DE+EB+BC

you should also add the equations DE=-FE and BC=-BA for clarity

then add them together in terms of their definitions

and do basic elimination

you don't need to visually represent the vectors

[delicate lance]

ohh ok

[dreamy sleet]

just put the little arrow above the letters to denote it

[delicate lance]

wait

to find points of inflection of an equation

do u do second derivative equals 0

[dreamy sleet]

yep

right on

[delicate lance]

wait so for question 14 i’m maximizing perimeter or what

[dreamy sleet]

the extrema of the function are the roots of the first derivative
the inflection points of the function are the extrema of the first derivative which are the roots of the second derivative

[delicate lance]

ohh ok

[dreamy sleet]

which one do you need help with

[delicate lance]

the word problems

wait

are u able to help me

or am i bothering u

cuz i can go to sleep

if u are busy

[dreamy sleet]

I'm here to help, It's the weekend so i have literally nothing else to do

I'll be up for another hour or so

[delicate lance]

😊😊😊😊

yay

[quasi saffron]

i think i need to learn myself from @dreamy sleet on how to explain well

hahaha

[dreamy sleet]

Just don't make any assumptions on what someone may or may not know, and never skip any steps

explain everything to it's entirety depending on what you know they know

[quasi saffron]

umm for a reference, where do you guys from? cuz it will be different on how we see things in this case is math

[delicate lance]

canada

[dreamy sleet]

I'm from Texas, currently in college and taking calcII as a summer course

q14: A rectangle has a perimeter of 100m. What length and width should it have so that its rea is a maximum. What is the maximum value of its area?

[delicate lance]

ohh

maximizing area

[dreamy sleet]

Yep

[delicate lance]

so do we do

[dreamy sleet]

first, define an equation for your perimeter

[delicate lance]

2l + 2w = 100

[dreamy sleet]

alright

now which variable do you want to use for the rest of it

[delicate lance]

i’ll do

[dreamy sleet]

w is normally easier to read

[delicate lance]

50 - l = w

[dreamy sleet]

I would recommend solving for l to replace it, just for the sake of good habits

but, if you'd like, we can move forward like this

[delicate lance]

lets move forward

[dreamy sleet]

alright

what's your area equation

[delicate lance]

so i plug in area

length time width

[dreamy sleet]

write it out

[delicate lance]

l(50-l)

= A

[dreamy sleet]

yeah, A(l)=l(50-l)

go ahead and distribute to make it easier

[delicate lance]

50l - l^2 = A(l)

so

[dreamy sleet]

alright, now take the derivative

[delicate lance]

50 - 2l = A’(l)

[dreamy sleet]

and solve for zero

[delicate lance]

l = 25

so is that just length

[dreamy sleet]

now that you have that, solve for width (spoiler: your final shape is a square)

[delicate lance]

50 - l = w

25

wait so

max area is

625?

[dreamy sleet]

25*25=625 yeah

[delicate lance]

im yayyaya

[dreamy sleet]

note that the rectangle with the largest area for the least perimeter will always be a square

i have it pulled up in my browser dw

[delicate lance]

wait

can we do trig limits

do u remember

them

[dreamy sleet]

Go ahead and show me a question and ill see

[delicate lance]

question uhhh

14

[dreamy sleet]

,tex $\lim_{\limits{x\rightarrow-3\pi}}x^3\sin^4(x)$

[solid kiln]

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

[dreamy sleet]

that works

anyways

[delicate lance]

long ahh arrow

don’t tell me u can just plug it in

i’ll be so disappointed

[dreamy sleet]

pretty sure you can

it would be 0

What do you want to cover next?

[delicate lance]

Hmm

Are u still here

I was looking at my instagram graduates page

[dreamy sleet]

yeah but I think i'm gonna hand off to another helper since i need to use the restroom

[delicate lance]

Oh okay

Thank you for your help

❤️❤️

[dreamy sleet]

Np

[delicate lance]

No its ok

I’m fine

I’ll go to bed

[wraith hinge]

Hola

[dreamy sleet]

Ah

I see

[delicate lance]

Sorry ache

[dreamy sleet]

I apolgise for the helpers ping

[wraith hinge]

Why

[delicate lance]

I don’t need help

I’ll close this channel

.,¥

[low maple]

wat

[delicate lance]

.close

[trim jolt]

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[wraith hinge]

:^

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[past quartz]

I am super confused, I have gone through all the steps for this problem but I dont understand what I have wrong

[hallow kite]

so you got from the inequality that 5+x >= 5 - x?

[vivid glade]

only if 5-x is positive

[hallow kite]

oh true, i guess you can consider cases

clearly x is never 5

cause of the denominator

so you can consider when x < 5 and x > 5, use that to determine the sign of the denominator, and then solve the resulting inequalities after multiplying over

[digital cloak]

A useful trick for these is not to multiply by (5 - x) but by (5 - x)^2

Cause that’s always nonnegative

[hallow kite]

the quadratic inequality is quicker to solve?

[digital cloak]

You don’t have to split into cases

[vivid glade]

that's a nice strat

if you're comfortable with factoring

[past quartz]

ill check it out, thanks guys

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[rapid horizon]

19x=1 mod 141

[rapid horizon]

I have solved it with euler division

Is there any other method? I want to learn all other aspects

[trim jolt]

@rapid horizon Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[marsh forum]

Simplify the following statements. Which variables are free and which are bound? If the statement has no free variables, say whether it is true or false.

$w\in\ \left{x\in\mathbb{R}\right|13-2x>c$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so here we input w , so it's a free variable

c is a bound variable as its a constant

[proper kernel]

can you explain what a bound variable is for me

[marsh forum]

A varaiable that constraints the set

it's what constructs the set

[proper kernel]

and what about a free variable

[marsh forum]

we input that to check if it's an element of the set or not

[proper kernel]

both of these definitions are not true

you can recheck: https://en.wikipedia.org/wiki/Free_variables_and_bound_variables

[marsh forum]

so here w would be a free variable and x a bound variable ?

[proper kernel]

w is free, x is bound, yes

what about c?

[marsh forum]

c is bound?

[marsh forum]

An instance of a variable symbol is bound, in contrast, if the value of that variable symbol has been bound to a specific value or range of values in the domain of discourse or universe.

[proper kernel]

you should consider scrolling down

to this image

what does this image tell you about the letter z?

[marsh forum]

z isn't constrained

so it'f free

[proper kernel]

z is a free variable right?

[marsh forum]

yes

[proper kernel]

so in ${x\in\mathbb R\mid13-2x>c}$, is $c$ free or bound?

[solid kiln]

mtt

[marsh forum]

free

[proper kernel]

there you go

[marsh forum]

oh,ok

thanks

[proper kernel]

np

[marsh forum]

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[earnest nymph]

hi im having trouble with this question

[earnest nymph]

this is the table, but i dont understand how its 0.1 probability and 0.3

[upper harbor]

.1 probability means 10% chance

[earnest nymph]

yea but idk how to get there

[whole sage]

@earnest nymph it is 0.1 as the probability as the segment 1 to 4 have a total probability of 0.4 and dividing that into 4 segments would give you 0.1. Hope that helps : )

[earnest nymph]

sorry, how is the total probability 0.4 from 1-4

Im not sure how to get there

[whole sage]

@earnest nymph sorry for my misunderstanding, the total probability is 0.4 as if you have a look at question where it says "Kurt, however is exceptionally good at this game and is 50% more likely to 5 or 6 than 1 to 4" and the probability of being 5-6 adds up to 0.6 which is 50% more than 0.4.

I can explain you more theoretically, than rather just guessing and checking, if so?

[earnest nymph]

So basically

If i assign prob of 1-4 as x and 5-6 as 1.5x and thats out of 1, i can solve for x?

[whole sage]

What are you trying to say by thats out of 1?

[earnest nymph]

Cause total probability will be out of 1

[whole sage]

Oh ok

[earnest nymph]

Is that acceptable

Cause it works

[whole sage]

Yes

[earnest nymph]

Oki thx

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[wraith hinge]

<@&268886789983436800>

[arctic quest]

why u @ mods?

[wraith hinge]

Someone was advertising

[cinder temple]

so true

[winged trench]

u want this?

[wraith hinge]

Yes

[winged trench]

did u try dividing p(x) by Q(x)?

to get value of b?

oi

@wraith hinge

[wraith hinge]

no

[winged trench]

u didnt try or its not easy?

[wraith hinge]

not easy

and i don't it will help

3 min

[vagrant prism]

what should a,b,c,d be if they are both in an arithmetic and in a harmonic progression

[wraith hinge]

i will try

[winged trench]

im getting a b c are in gp as well

[wraith hinge]

1's ?

[vagrant prism]

can't all be ones, since a<b<c<d

[winged trench]

am i the only one?

[vagrant prism]

i didn't test that

[wraith hinge]

missed that condition

[winged trench]

i equated the ap and hp condition

and got them to be in gp

[vagrant prism]

hm i don't see it

[winged trench]

what is the hp condition

1/a + 1/c = 2/b

ryT?

[vagrant prism]

ooh in that sense

[winged trench]

from ap i got a + c = 2b

and putting it here

i got ac=b^2

[vagrant prism]

yep

interesting

[wraith hinge]

very

[winged trench]

am i wrong or am i correct?

and is there smtg u can do with that info?

[wraith hinge]

4 numbers which are in A.P , G.P and H.P
Is there a particular set which satifies this?

[winged trench]

-_-
dont tell me ur brainstorming this

[wraith hinge]

?

[thorny ore]

hello

[winged trench]

not sure if the 4 r in gp

the 3 r tho for surr

e

[winged trench]

hello there dark knight

[thorny ore]

hiya

from what i can see you can use sum and product of roots of cubic equation

that shall give you the value of a,c,d

[winged trench]

hmm lemme try

[wraith hinge]

2,3,4 but they wouldn't satisfy the third case i.e sum of product's of pair wise roots

[winged trench]

uh how did u get this?

[wraith hinge]

24 = 6* 4 = 2 * 3 * 4

9=2+3+4

hmm

[thorny ore]

no bruh

use the formulas of sum and products of roots

[wraith hinge]

solution says 3 of the roots are 2,3,4

[wraith hinge]

thats what i did

24= product of roots

9 is the sum

[winged trench]

uhh this is what i got

2b + d = 9
d*b^2 = 24

and

b^2 +2bd + a = 0

[wraith hinge]

wrong question ?

[winged trench]

no the roots r not in ap?

they missed b

[thorny ore]

yea the question seems to be wrong, 'b' shouldn't be in AP

it should be only a,c,d i think

[winged trench]

hmm

[wraith hinge]

OKay

thanks!

.close

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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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.

[mint pier]

Hm

[thorny ore]

any question?

[mint pier]

Nope

[thorny ore]

bruh

stop playing

.close

[marsh forum]

can I close this or do you have anything to ask

[thorny ore]

yea dude's just playing

close it pls

[mint pier]

Waait let me type the question 😈

Let start with a simple algebraic equations

[marsh forum]

,rotate

[mint pier]

<@&286206848099549185>

[marsh forum]

!15m

[trim jolt]

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.

[marsh forum]

anyway, substitute x=-bc/ad in the equation

and you'll have your answer

[mint pier]

I know that much i m just testing ur skills

Give me the solution with each step

👀

@marsh forum

[marsh forum]

!nosols

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[marsh forum]

I won't give out answers

[mint pier]

Bro im graduate

[marsh forum]

stop trolling then

[mint pier]

I just want to test everyone here that how skilled they are

[mint pier]

Bro it only u who thinks that he is getting trolled 👀

@everyone

Look @marsh forum cant solve a simple question

[winged trench]

smh

[oblique valley]

People active in this Discord know he can

[winged trench]

just close the channel if you know u can solve

[oblique valley]

Could you please stop trolling and let the channels to people who actually need help?

[mint pier]

I just wanted to test his skills but after all , I think he failed 10 grade

[marsh forum]

<@&268886789983436800> troll

[winged trench]

well now that ur done testing
just close the channel

[mint pier]

I won't

[winged trench]

great

[dull temple]

unlucky

.close

[trim jolt]

Channel closed

Closed by @dull temple

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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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.

[wraith hinge]

This is what i did
height of equilateral triangle = $sqrt(3)/2* a^2$

[solid kiln]

Cnidarian

[winged trench]

failed miserably

[wraith hinge]

this is what i got

[wraith hinge]

well,not that bad
i got to some value

!status

[trim jolt]

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

[wraith hinge]

4

[winged trench]

the command part i meant

the farther vertex would mean -ea,0
im bad at hyperbola so just confirming

[wraith hinge]

nope,its (-a,0)

[winged trench]

oh mb its vertex not focus

the latus rectum is the opp of the focus ryt

?

[wraith hinge]

?

[dull temple]

the what

[winged trench]

ugh

i meant

[dull temple]

oh that's not a typo that's just weird Greek

[winged trench]

the taken triangle is of the latus rectum of one side and vertex of the opp side ryt?

[winged trench]

ye i tend to do that

[winged trench]

ha ok so this

[wraith hinge]

yes

[winged trench]

so

root5/2 b^2/a = a(1+e)

ryt?

[wraith hinge]

3

[winged trench]

wait y root3

[wraith hinge]

what the height of a equilateral triangle if the side length is a ?

[winged trench]

tf

wait

oof my bad

i added instead of subtracting

ye 3

[winged trench]

what do u get from this?

and e = a^2 + b^2 by a^2

[wraith hinge]

^

[winged trench]

u sure?

thats messed up

[wraith hinge]

yes

[winged trench]

im getting diff?

[wraith hinge]

what is it

[winged trench]

root2 / root 3 -1

@wraith hinge

im getting diff e value

(root3 +1) / root 2

[wraith hinge]

how are you getting a root 2

[winged trench]

i got e^2 = root3 +1 /root3 - 1

[winged trench]

from lierally this

[wraith hinge]

answe

[winged trench]

oof

[wraith hinge]

oh

f

[winged trench]

what?

[wraith hinge]

e^2

wrong solution,wrong question

[winged trench]

smh

[wraith hinge]

sorry!

.close

[trim jolt]

Channel closed

Closed by @sonic oasis

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[marsh forum]

$\lim_{x\rightarrow0^+}\left[\sin\left(x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin\left(x\right)}\right]$

[trim jolt]

Please don't occupy multiple help channels.

[marsh forum]

oops

closed the other channel

so this is of the form $0^\infty$+ $\infty^0$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so I evaluvate each limit using logs

right

and then add them up

[lapis star]

is the ans 1?

[marsh forum]

yes

LH is one way I guess

but I'd rather not use it

[main sigil]

Is that limit from right side?

0+?

[marsh forum]

oh right

yeah

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

so I guess I can approximate sin(x) to be x

series expansion I guess

[tulip vapor]

Damn @marsh forum you're soo active here

[marsh forum]

yup, it keeps me sane

[lapis star]

ye banda kbhi nhin sota

[tulip vapor]

Bro tere koi exams nhi hai kya

[marsh forum]

over, I'm starting uni this year

[main sigil]

Well technically

[tulip vapor]

Which college

[main sigil]

more formally, rewrite it as e^something

[shy nacelle]

sinx^(1/x) = e^(ln(sinx)/x))

[main sigil]

yup, this

[winged trench]

banda means monkey?

[marsh forum]

I don't know

[amber python]

or just say sin(x) = x and yolo

[pulsar dust]

No

Bandar

[winged trench]

ha exactly

[main sigil]

fundamental theorem of engineering, my favorite

[pulsar dust]

Means monkey

[lapis star]

cool

[pulsar dust]

FTE comes in clutch

[lapis star]

@marsh forum u gave jee?

[marsh forum]

I thought of using that , but that just gives me $x^{\frac{1}{x}}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[vagrant prism]

you can deal with that

[marsh forum]

true

let me try

[lapis star]

rank?

[shy nacelle]

f(x)^g(x) = e^[(f(x)-1)*g(x)]

[pulsar dust]

Slight mistake if it's close enough then it's equal

[winged trench]

never used pi = e

[main sigil]

💀

[marsh forum]

$\frac{1}{x}\ln\left(x\right)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[winged trench]

hmm jee folks here?

[marsh forum]

this is ln(L)

[main sigil]

Isn't that too much

[shy nacelle]

No

[main sigil]

you are over-engineering it

[marsh forum]

this is wrong?

[shy nacelle]

I learnt it JEE

[main sigil]

x is approximately 0 though

[shy nacelle]

Im not egyptian

[main sigil]

ln(x) = x-1 works for x around 1

[vagrant prism]

eh, not gonna help much

[winged trench]

i didnt expect lol

[vagrant prism]

keep it as e^(1/x ln(x))

[main sigil]

(unless you are an engineer)...

[lapis star]

anyways is the answer 1?

[vagrant prism]

yes

[marsh forum]

yes

[lapis star]

oh i got it

[vagrant prism]

why? ln(x) just drops REALLY quickly compared to how 1/x rises

so 1/x ln(x) will go to -inf

and e^-inf is 0

[marsh forum]

ooh

right

makes sense

[vagrant prism]

also

[main sigil]

im bit lost at the moment

[vagrant prism]

ln(x) is negative for x<1

and 1/x is still positive

[main sigil]

oh ic nvm

[lapis star]

i used simple logic

[vagrant prism]

same

[pulsar dust]

Same

[shy nacelle]

Why didnt you use the e^form bro

It gets solved easily that way no?

[marsh forum]

that's only for $1^{\infty}$no?

[main sigil]

you cant use that f(x)-1 thing

[solid kiln]

ƒ(Why am. I here)=I don't Know

[main sigil]

ln(x) = x - 1 works for x near 1

[vagrant prism]

not necessarily

[main sigil]

but here x approaches 0

[pulsar dust]

If it works it works

[shy nacelle]

No not that

[lapis star]

sin(0) ^ inf will be 0 and the second term (inf)^0 will be 1, my one line compact soln xD

[shy nacelle]

Just raising to e^ power and l hopital

[winged trench]

l hospiatal everything and anything out of existence

[shy nacelle]

e^( sinx ln 1/x )

[marsh forum]

I'm doing this so that I'll find uni easy, I would like rigour lol

[lapis star]

which uni?

[winged trench]

hm uni maths like this?

[marsh forum]

My mom said, I know it's not, just have to keep her happy

[shy nacelle]

Or e^-(ln x/cosecx)

[shy nacelle]

Which class are you?

[pulsar dust]

I see i need to do the same then , but my lazy probably won't

[main sigil]

what

[marsh forum]

A decent pvt uni for maths

[lapis star]

cool

[shy nacelle]

Manipulation

Now use L hopitals

[winged trench]

lol true ig

i wish there were no reservations

[marsh forum]

this channel is gtting off topic fast

[winged trench]

true

[shy nacelle]

…

[marsh forum]

huh

[shy nacelle]

after l hopital

[marsh forum]

yeah

that works

[shy nacelle]

Then ?

[main sigil]

Whattt

do you realize that its not the same limit

like at all (nvm it sorta is but the manipulation was probably random af)

im surrounded by engineers

[tulip vapor]

@marsh forum bro aapne jee diya na

[marsh forum]

yes, but please keep the channel on topic

[tulip vapor]

Rank?

[frozen plover]

I'm enginearing my fucking limit

[shy nacelle]

Why?

[main sigil]

how did the sine get outside of ln

[pulsar dust]

Yeah

[shy nacelle]

The first part will be 0 isnt that obvious

[shy nacelle]

Im speaking of 2nd part

[main sigil]

oh

[marsh forum]

$e^{-\frac{1}{x\operatorname{cosec}\left(x\right)\cot\left(x\right)}}$

[pulsar dust]

Oh

[solid kiln]

ƒ(Why am. I here)=I don't Know

[main sigil]

why would you write it as /cosec though

that's insane

[marsh forum]

so the limit is 1 again

[shy nacelle]

For first part we get e^[ln(sin x)/x]

[marsh forum]

yes

now we use LH in the numerator

[main sigil]

not again though

[marsh forum]

I mean the exponent

[shy nacelle]

That is -∞/0+

e^(-∞/0+)

[marsh forum]

which is 1

[tulip vapor]

Why are y'all discussing it over and over if you know the answer is 1 💀

[marsh forum]

the method matters more than the answer

[shy nacelle]

We need process

[tulip vapor]

Why tho

I thought steps only mattered till school

[amber python]

blud hasnt done any serious maths

and it shows

[tulip vapor]

You're right, that's why I'm asking lop

Lol

[shy nacelle]

Bro broke the matrix

[pulsar dust]

It matters even more after school

[amber python]

you will never get anywhere if you dont understand how to do anything

[amber python]

people are not machines

[pulsar dust]

University

[marsh forum]

research ?

[amber python]

wolfram can compute the limit in less than a second

[frozen plover]

snow preaching her pure math propaganda

[amber python]

not even pure math smh

[marsh forum]

ok, we're seriously digressing

[tulip vapor]

So like if you got the answer in your head, why think of it over and write down the steps

[amber python]

if you know how to get the answer, then all the more reason you should be able to write it down!

[pulsar dust]

Think it like this mathematicians have trust issues

[amber python]

if you are unable to write it down, then it means you never understood in the first place

[tulip vapor]

So did the original poster have trouble writing their solution

[marsh forum]

I didn;t even know where to start lol

I mean logs came to mind

but I didn't see how that would help

becuase it would give me 0/0 again

[main sigil]

whenever it's a fraction, you can at worst apply lhop and smh solve it

[frozen plover]

you wanna compute
[ \lim_{x \to 0} \f{\ln \sin x}{x} ] without LH right?

[pulsar dust]

For which did you get it

[solid kiln]

Bob l'éponge

[marsh forum]

Not going to work at uni

[frozen plover]

no wait

the other one?

[marsh forum]

the first part

[main sigil]

it was -ln(x)*sin(x) I think?

or sth like that

[marsh forum]

let's start this afresh ?

[main sigil]

the second part?

[pulsar dust]

So this w/o lh?

[marsh forum]

yea

[frozen plover]

LH doesn't really apply there

[marsh forum]

$\lim_x\rightarrow_{0} (\frac{1}{x})^{sin(x)}$

[main sigil]

{}

you need to group them i believe

[solid kiln]

ƒ(Why am. I here)=I don't Know

[pulsar dust]

$\lim_{x\rightarrow_{0}} (\frac{1}{x})^{sin(x)}$

[solid kiln]

convergence

[frozen plover]

[ \lim_{x \to 0} \left ( \f 1x \right )^{\sin x} ]

[marsh forum]

yes

[solid kiln]

Bob l'éponge

[frozen plover]

my latex best latex

[marsh forum]

now taking the log on both sides

$ln(L)= sin(x) ln(((1/x))$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[main sigil]

limit of that ofc

sin(x)ln(1/x)

x->0+

[marsh forum]

that would be 0

so L=1

[frozen plover]

why?

[marsh forum]

$ln(1/x)--->-\infty$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[frozen plover]

okay

[main sigil]

-infinity * 0 is indeterminate

[pulsar dust]

$ln(L)=-\sin(x)ln(x)$

[solid kiln]

convergence

[frozen plover]

also positive infinity right

[marsh forum]

right

my bad

so doesn't work here

*here

[frozen plover]

how about x = e^t does that help

x --> 0 => t --> -infinity

[marsh forum]

hmm

[main sigil]

maybe just multiply by x/x first?

to make the sin go away

[pulsar dust]

Yes

[frozen plover]

smart

[main sigil]

i actually got this from someone else

[frozen plover]

yeah okay obvious then (what is to be done next)

[main sigil]

whenever you see sin, it can be replaced by multiplying it by x/x

[pulsar dust]

Not snart

[main sigil]

or whatever is in the argument

[marsh forum]

so I have to find the limit of $-xln(x)$ at $x=0+$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[frozen plover]

or taylor polynomial it to just x

[main sigil]

Yes

[pulsar dust]

Skill issue

[frozen plover]

polynomial stronger than logarithm

[pulsar dust]

Anyhow back to the question

[marsh forum]

that works, but will it fly at uni

I doubt it

[frozen plover]

wdym

[marsh forum]

have to series ln(1/x^x) I guess

[frozen plover]

what

[main sigil]

isnt xln(x) one of the limits to remember?

[frozen plover]

why

[main sigil]

like sin(x) / x

[marsh forum]

I didn't know that

[pulsar dust]

I think he needs more rigor

[frozen plover]

you're fine with exp > poly then why not poly > log

[frozen plover]

fine whatever

sub x = e^t

surely you can use exp > poly

just saves time

[main sigil]

note that then t -> -infinity

[marsh forum]

wait, what

$-e^tln(e^t)$

?

[solid kiln]

ƒ(Why am. I here)=I don't Know

[main sigil]

ln(e^t) = t

[marsh forum]

ues

*yes

ah

makes sense

thanks

[pulsar dust]

@marsh forum it doesn't need that much rigour state it as trivial

[frozen plover]

and why did that not work before the sub

[marsh forum]

Will I even come across such limits at uni?

or is this limited to pre-uni

[pulsar dust]

1 year maybe

[marsh forum]

. Thanks

[pulsar dust]

It's not that rigorous

[frozen plover]

tbf I doubt you'll be computing limits like this

[pulsar dust]

But analysis will be

[frozen plover]

you'll have to prove it

[main sigil]

is this limit even solvable? We need to have at least some "known" limit to work it, dont we?

like xln(x)

or xe^x

[marsh forum]

yeah

[twilit latch]

Epsilon delta is going to be be your life in early analysis

[main sigil]

or epsilon delta

[marsh forum]

I know, but such limits ?

[amber python]

you would prove the limits involving x log(x) or x e^x

thats what an analysis class is for

[marsh forum]

Okay, got it, thanks

[pulsar dust]

Not really unless your prof hates your class to the core

[amber python]

not as an exercise

but at some point during the course it should be proven

unless your course sucks

then maybe not

[twilit latch]

Also a lot of series stuff with terms like this

Which can be worse

[marsh forum]

ah, hopefully this will make more sense once I understand series

thanks

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[marsh forum]

thanks everyone!

[amber python]

this doesnt have to use series

[pulsar dust]

Welcome if I helped you

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[main sigil]

.that was quick

[twilit latch]

I meant just the natural progression to limit on sequences is limits on partial sums for series in analysis

[main sigil]

Hello, what's your question?

[twilit latch]

.close

[trim jolt]

Channel closed

Closed by @twilit latch

Use .reopen if this was a mistake.

[amber python]

and im trying to make sure that @marsh forum doesnt get the misconception that this question requires knowedge of series to prove

[marsh forum]

got it

thanks

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[tulip vapor]

(tan^2 pie/7 + tan^2 2pie/7 + tan^2 3pie/7) * (cot^2 pie/7 + cot^2 2pie/7 + cot^2 3pie/7)

[tulip vapor]

<@&286206848099549185>

[cinder temple]

wat

[cinder temple]

yea shouldve ping after 15 mins

[exotic pine]

$(\frac{\tan^2{\pi}}{7}+\frac{\tan^2{2\pi}}{7}+\frac{\tan^2{3\pi}}{7})(\frac{\cot^2{\pi}}{7}\frac{\cot^2{2\pi}}{7}\frac{\cot^2{3\pi}}{7}$)

this??

[solid kiln]

Skill_Issue

[wraith hinge]

ate

[vivid glade]

me when parentheses don't match size of contents 😔

[wraith hinge]

skill issue indeed 😔

[junior saffron]

you forgot the + signs in the second parethesis expression

[cinder temple]

u forgot to add plus sign before cot^2pi/7

[wraith hinge]

on it gang

$(\frac{\tan^2{\pi}}{7}+\frac{\tan^2{2\pi}}{7}+\frac{\tan^2{3\pi}}{7})+(\frac{\cot^2{\pi}}{7}\frac{\cot^2{2\pi}}{7}\frac{\cot^2{3\pi}}{7}$)

[rugged latch]

@tulip vapor are pi/7, 2pi/7… arguments of the functions or is it just pi, 2pi

[solid kiln]

mid

[vivid glade]

mmmm this is just too much. this is all wrong

[cinder temple]

thats like (cot^2pi/7)'s cube power

[junior saffron]

you mean $$(\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}) \cdot (\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7})$$

[solid kiln]

Crystopher

[wraith hinge]

not the danimals profile 😭

[cinder temple]

thats right

[wraith hinge]

skill issue came up with the latex, I just added a + sign in the second parenthesis expression 😭

[vivid glade]

$$\left(\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}\right) \cdot \left(\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}\right)$$
great work team

[solid kiln]

Flip

[vivid glade]

problem solved, .close

[verbal idol]

wow

[wraith hinge]

bro aryan hasn't even commented yet lmaoo

[tulip vapor]

im back guys

[verbal idol]

alr

[cinder temple]

tbh this quetsion's calculation amount is pretty big

[tulip vapor]

let me go thru what yall have written

[verbal idol]

ig you can write tan^2(3p/7) = tan^2(4p/7) and that should help

[tulip vapor]

arguments

[tulip vapor]

how lol

[cinder temple]

i wonder where is that math problem from

[tulip vapor]

$$\left(\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}\right) \cdot \left(\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}\right)$$

[solid kiln]

Aryan07

[tulip vapor]

yeah this is the one

[tulip vapor]

like 11th grade

we need to find numerical value

how do i solve

[verbal idol]

what have you tried yet?

[cinder temple]

rly crazy

[tulip vapor]

i dont know what to begin with

[verbal idol]

Indian ho na? hindi chalegi?

[cinder temple]

i thought that was like university

[tulip vapor]

ha

[verbal idol]

han toh pi/7 ko theta man lo

[tulip vapor]

ok

[verbal idol]

toh 7A = pi

[tulip vapor]

hmm

[cinder temple]

i dont feel well im out

[verbal idol]

wait brb

[tulip vapor]

bruh

can we use pie/7 as theta and then use single, double and triple angle identities

but that would be for tan only, then we need to square it

[verbal idol]

bahut lamba jayga

[frozen plover]

sounds like death

[tulip vapor]

yeah

[verbal idol]

yeah

[dense breach]

I would start by either just distributing or using a relation between tan and cot

[tulip vapor]

whats the other method then

[pulsar dust]

No it is death

[verbal idol]

@tulip vapor topper pe hai solution you can check it ig

[wet dragon]

its pi not pie

[wraith hinge]

byee pluto

[verbal idol]

https://www.toppr.com/ask/question/find-the-sum-of-digits-of-the-numerical-value-of-lefttan2fracpi7tan2frac2pi7tan2frac3pi7rightleftcot2fracpi7-cot2frac2pi7cot2frac3pi7right/

yeah its available online

[wet dragon]

just expand

[tulip vapor]

bruh i looked at solution it gives me headache

[verbal idol]

yeah avg question from compound angles / tri if you're preparing for jee

[tulip vapor]

they used some wierd 4theta formula i dont even know

[verbal idol]

oh, you can derive it

[tulip vapor]

yeah but then it will be 2theta + 2theta and it will be too long

[verbal idol]

theres are general formula for that

you won't learn that in school tho

[tulip vapor]

in coaching too they only taught till triple angle

[verbal idol]

check module

it might be there

[tulip vapor]

it isnt there

[wet dragon]

(a+b+c)(d+e+f)= ad+af+bd+bf+cd+cf+ea+eb+ec

just expand lolz

[tulip vapor]

bruh

[dense breach]

No I mean nobody is gonna look at this and think quadruple angle

are there any findable solutions

[verbal idol]

yeah you have to practice these types

[tulip vapor]

as a note maximum time you can give to this question is like 5mins

[frozen plover]

tbf this isn't a bad idea cuz a bunch of it will cancel out

[wet dragon]

beta

https://www.doubtnut.com/qna/8492746

[verbal idol]

there is a formula for tan (nx)

[frozen plover]

and you have patterns to work with

[wet dragon]

see if this video helps

nvm its the smae thing

[molten maple]

what is the question?

[tulip vapor]

its exact same thing as topper solution

$$\left(\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}\right) \cdot \left(\cot^2 \frac{\pi}{7} + \cot^2 \frac{2\pi}{7} + \cot^2 \frac{3\pi}{7}\right)$$

[wet dragon]

ai8ght i will do a challenge

[solid kiln]

Aryan07

[wet dragon]

i will expand it

and let the video play

lets see who is faster

[tulip vapor]

do it in this chat

bruh i need a simpler solution f

r

this isnt even some jee main or advance question

[dense breach]

you get lots of ratios

tan(2π/7)/tan(π/7)

I think those simplify a bit

using double angle

[frozen plover]

i dont like this question

[tulip vapor]

neither do i

but gotta solve thi

this

guys

[verbal idol]

bro

the solution

is the best

[tulip vapor]

lets do teamwork, 3 of us take one term each from LHS and multiple with RHS

then combine all 3 answers