help-23

[lean otter]

So i think so

[frozen marlin]

yes

it won't decrease though

it'll be zero

the SECOND derivative will be zero--the derivative will b decreasing

but the function won't immediately start decreasing

[lean otter]

🤔

[frozen marlin]

do you understand?

[lean otter]

sry i don't

[frozen marlin]

at the maximum it STABILIZES
it doesn't immediately start decreasing

[lean otter]

so it will be a straight horizontal line for a while before decreasing?

[frozen marlin]

yes

for an infinitesimally small portion

[lean otter]

I see

[frozen marlin]

otherwise there would b a very sharp "break"

[lean otter]

i need to find that point?

[frozen marlin]

yes

BUT

derivatives are zeroes at two points

minimums AND maximums

find these points, plug them into the function, and take whatever the maximum is

[dire fjord]

note: f'(a)=0 isnt always the same as (a,f(a)) is a maximum/minimum of the curve of f

but the other way is true

[frozen marlin]

i know

but for this graph it isn't constant

*a constant

so won't really matter, it involves exponentials anywya

[dire fjord]

so if a point is a max/min means that f'=0 but the other way isnt always true

[frozen marlin]

*anyway

[frozen marlin]

ik but doesn't apply here

[dire fjord]

so you were talking about this case i noted that bc you sounded general

thats all

[frozen marlin]

oh alr

[lean otter]

So can I just write the derivative of the function and equal it to 0 to get the answer?'

[frozen marlin]

yes

[lean otter]

thanks sm bro ur a life saver

ill hyu when i get an answer

C'(x) = 3.6x + 54 - 0.2x^1.5 = 0

3.6x - 0.2x^1.5 = -54

problems is there are 2 x's idk how to go on from here

[frozen marlin]

sorry im back

ummm yea that's right

i think just plug it into WA? idk how to solve such equations

[lean otter]

no worries

whats WA?

[frozen marlin]

wolframalpha

or just use a calculator

[lean otter]

i feel like somethings wrong

[frozen marlin]

??

wdym

dude look

nothing's wrong

this is standard--and it had a decimal power anyway

it would have been hard to solve without the derivative

okay

anyway i plugged the equation into desmos b4, i got the same value of 'x' from wolframalpha

i think you're good to go

yea

ur pretty much good to go!

[lean otter]

Oh mb

[frozen marlin]

n[

*np

i gtg now, lmk if u need anything else

sorry, i hope i was of help

gn

[lean otter]

gn bro thanks for the help again

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[limpid karma]

I get no real solutions

[limpid karma]

I used integration

and got -4ln(cos(b)) = 2nln2

but this has no real soluitons

[woven hound]

I'm not sure you integrated correctly

wait nevermind

$-4\ln{\cos{\phi}}=2\ln{2}$

[flat frigate]

Ari

[woven hound]

$\ln{\cos{\phi}}=-\frac{1}{2}\ln{2}$

[flat frigate]

Ari

[woven hound]

$\ln{\cos{\phi}}=\ln{\frac{1}{\sqrt{2}}}$

[flat frigate]

Ari

[woven hound]

$\cos{\phi}=\frac{1}{\sqrt{2}} \implies \phi = \frac{\pi}{4}$

[flat frigate]

Ari

[woven hound]

uh but we're solving for b

oh so b is pi/4

[limpid karma]

ohhhhhh i see my mistake now

thank you sm!

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[covert abyss]

Could someone help me with this question

[ancient escarp]

have you attempted anything yet?

[covert abyss]

yes kinda but it was wrong

[ancient escarp]

can you show what you tried?

[lean otter]

Can you convert
h² + 3h - 70 into simpler forms using splitting the middle term of something?

[covert abyss]

i cant since i wrote it on paper and im on a pc

[covert abyss]

not sure

[ancient escarp]

okay. can you explain what you tried?

[covert abyss]

i really dont know how to

[ancient escarp]

have you learned how to factor?

[covert abyss]

its fine thank you for trying to help

[covert abyss]

yes

[ancient escarp]

okay, so you see that quadratic on the denominator to the right

do you think you could factor that?

[covert abyss]

yeah its fine i think i get it a bit

[lean otter]

How would you factor
h² + 3h -70 ?

Like what do you know

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[eternal creek]

Where does this come from the q^c u d^c

Is it because q is defined as q intersect d?

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@eternal creek Has your question been resolved?

[regal lodge]

d^c n d is empty set so it doesn't affect the equation so adding it to the left doesn't change it
(q^c n d)u(d^c n d)=(q^c u d^c) n d

[eternal creek]

Alright thank you

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[lean otter]

this is more of a general question

[lean otter]

if a hyperbola is parametrised by (a cosh t, b sinh t) where a> 0 and cosh t > 0

how does the hyperbola exist in negative x co ordinates then ?

cz wouldnt a cosh t always be > 0 ?

[median vigil]

that doesn't by itself parameterize the entire hyperbola, only the branch with positive x coordinates as you noted. but if you take (-a cosh t , b sinh t) with the same a, b as before then that would parameterize the other branch

[lean otter]

ohh

tyty

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[robust veldt]

can someone help me with this question. i have to find out what to put instead the 0, all the other solutions are correct

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[lilac vine]

Can someone help me with Part C I don’t know where to start the other parts are correct

[plucky elk]

are you taking the ap exam?

[lilac vine]

Yes in may

This is a practice my teacher gave for the quiz we have

[plucky elk]

semi-circle cross sections implies something is being rotated.

[lilac vine]

Ok

What do you do after

[plucky elk]

did you learn any equations involving rotation

[lilac vine]

We learned volume and area but not cross sections

[plucky elk]

volume is made up of cross sections

https://tutorial.math.lamar.edu/classes/calci/volumewithrings.aspx

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[west halo]

Can someone help me do b(i)

[shadow glade]

i think probably draw a diameter\radius like that, then you can form a right triangle with the radius and a segment to the edge of the hexagon and calculate based on the dimensions of the pencil stuff that i assume you got from previous part

[west halo]

Can you elaborate a bit more

[shadow glade]

what info do you have about the hexagons \ circles inside the pencils at this point?

you know their various dimensions i assume?

the hypotenuse of that is the radius of the outer circle

the long leg of it is going through the center of both small circles and then to the midpoint of the hexagon's outer line segment

[digital bay]

t

t

tttretyutttttytr

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[west halo]

Oh yes thanks got it

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[patent vault]

how would i do this problem?

[eternal carbon]

(what have you tried so far?)

to start, you'd clear denominators (i.e. multiply both sides by pxy)

[patent vault]

oh ok

@eternal carbon i get
2xy=py+px
2xy=p(x+y)

does this show that there is only one solution?

[empty gyro]

consider p>2 a prime a nd what that means in relation to 2xy

[patent vault]

oh ok

[patent vault]

the number might be divisible by p?

you there?

@empty gyro

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[patent vault]

@Heleprs

<@&286206848099549185>

anyone?

[glacial mesa]

Hi

hmm

well poop idk this

What subject of math is this

[patent vault]

proofs and logic

[glacial mesa]

What class is this

[patent vault]

its from a extracurricular i do

[glacial mesa]

Oh

Do you know grade level

Looks like uni level at first look

Hi

Hmm

[patent vault]

High school

[patent vault]

Anyone?

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[mortal ore]

I have these two equations. I need to find when there will be no solution.

6x - 5y = 3
3x + ay = 1

I know that there will be no solution when the lines are parallel with different y-intercepts

[mortal ore]

But I don't know how to solve for a

[stray socket]

There will be no solution if the coefficient of y/coefficient of x of both equations are the same and both equations are not the same exact line

[mortal ore]

a = 5/2

[stray socket]

Close

The coefficient of y in the first equation is -5

[mortal ore]

ah, so -5/2

[odd pelican]

No solution if
6/3=-5/a≠3/1

[stray socket]

Yup

[mortal ore]

thank you

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[sour token]

is this wrong?

i got 0,6,6,6,6,6

because -3x2 is 6 no?

ohhhhh

nvm

im good

im dumb lol

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[left orchid]

i dont know how to graph this or find the Y

[odd pelican]

y=mx+b

If lines perpendicular product of slopes -1

[left orchid]

let me try

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[lean otter]

Why do calculators keep giving me 19.2? What am I getting wrong?

[quasi bison]

multiplying 80 * 0.24?

you're not doing anything wrong.

what makes you say you are?

@lean otter

[lean otter]

every calculator I use said that the answer is 19.2

not 19.20

[quasi bison]

do you think 19.2 and 19.20 aren't the same number

[lean otter]

isnt 20 a bigger number than 2(sorry i have a learning disability)

[quasi bison]

20 is bigger than 2 but neither 20 nor 2 themselves appear here

do you know how decimals work?

[lean otter]

im learning about it now

[quasi bison]

0.2 = 2/10

and 0.20 = 20/100

these fractions are equal

[lean otter]

i kind of get it...i will just keep doing more questions until i eventually get it right thanks

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[random spindle]

Is this correct

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[plucky elk]

Most of this is unreadable

Flipping fractions for the sake of it is also quite nonsensical

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[umbral comet]

I used the quadratic formula to solve u^2 + u - 12 and for some reason my solutions were -4 and 3 why is this? I was able to use -3 and 4 to solve the quadratic equation (u-3)(u+4) how am I thinking of this wrong

[peak estuary]

what exactly is the issue?

if x is a root, then (u-x) is a factor

because if you plug u=x in, that term becomes zero

so if you plug u=3 into (u-3)(u+4) then the first factor is zero

and if you plug u=-4 in the second factor is zero

[umbral comet]

I need to find numbers that result in u^2 + u - 12

I guess -4 is valid

Now that I input it into u etc.

But it doesn't help me find the number needed to get the above quadratic equation u know what I mean?

As a factored polynomial if I'm saying that right?

[peak estuary]

no I dont know what you mean

you got 3 and -4 from quadratic formula

and then from those you can read off that (u-3)(u-(-4))=(u-3)(u+4) is the factored form

which is exactly what you wanted

or otherwise, if you are given (u-3)(u+4) then you can read off that 3 and -4 are the roots

if you multiply out (u-3)(u+4) you get u^2+4u-3u-12=u^2+u-12

as expected

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[umbral comet]

Yeah I see what you mean thanks

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[humble nexus]

a questions realted to the n-queens problem, the question says that we can answer in this way, but i dont quite get what this formula means (does x have to be always lower than y ? or) here is my interpretation

[humble nexus]

BTW in D x,y x represents x-axis , y in the same way is y-axis

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[forest plaza]

Point 𝐷 is the midpoint of bisector 𝐵𝐿 of triangle 𝐴𝐵𝐶. On the segments 𝐴𝐷 and 𝐷𝐶 we find points 𝐸 and 𝐹 respectively such that ∠𝐵𝐸𝐶 = ∠𝐵𝐹𝐴 = 90∘. Prove that
the points 𝐴, 𝐸, 𝐹, 𝐶 lie on the same circle.

[ruby bolt]

<@&268886789983436800>

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[forest plaza]

<@&286206848099549185>

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[forest plaza]

<@&286206848099549185>

<@&286206848099549185>

[golden forge]

to help visualizing it

sorry cuz idk the answer either

[rose holly]

Yes

[grim plover]

If 4 points lie on a circle, then the quadrilateral formed by them must be cyclic

If it's cyclic, opposite angles sum to 180 deg

[forest plaza]

i think that power of point help solve that. but i dont know how to explain that CE, AF and BH intersect at the same point

also BH is height of triangle from B to AC

[grim plover]

Prob intended method try this @forest plaza

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[forest plaza]

Point 𝐷 is the midpoint of bisector 𝐵𝐿 of triangle 𝐴𝐵𝐶. On the segments 𝐴𝐷 and 𝐷𝐶 we find points 𝐸 and 𝐹 respectively such that ∠𝐵𝐸𝐶 = ∠𝐵𝐹𝐴 = 90∘. Prove that
the points 𝐴, 𝐸, 𝐹, 𝐶 lie on the same circle.

[safe radish]

@forest plaza Has your question been resolved?

[quasi yarrow]

can you please give drawing?

[forest plaza]

@quasi yarrow

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@forest plaza Has your question been resolved?

[eternal carbon]

what have you tried / proven so far?

[karmic hedge]

this is hard to do synthetically without guessing the circle

so try to guess the circle first with a good diagram

[forest plaza]

Nope. But i think that power of point help solve that. but i dont know how to explain that CE, AF and BH intersect at the same point

also BH is height of triangle from B to AC

[eternal carbon]

you could start with this i suppose

using the height BH, what concyclic points can we identify in the diagram?

[forest plaza]

BEHC and BFHA

[eternal carbon]

yeah that's right

at this point it might help to have a diagram where AC is oriented on the horizontal

[eternal carbon]

wondering have you solved this?

[karmic hedge]

i realized that i made a mistake, but i have an idea

[eternal carbon]

from here, because BD is the angle bisector, i'd probably draw ||perpendicular line to BD passing through B||

[eternal carbon]

👀

[karmic hedge]

i have a different diagram that ill post
||basically, guess that hte circle is (BIC), let (AHC) intersect (BIC) at E, and then (AEB) will be tangent to AI (this is the part that i have to work out). then it suffices to show that D is on the radical axis of (AEB) and (BIC) which can be done wiht linearity of power of a point)||

||I is the incenter||

[eternal carbon]

[karmic hedge]

||this proves E is on BD, and then you can do the same thing and show F is on BD||

ill try to find a proof that ||(AEB) is tangent to AI||

[safe radish]

@forest plaza Has your question been resolved?

[safe radish]

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[safe radish]

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[tight void]

@placid oak !!

[tight void]

how are you doing

[placid oak]

uhh, ok

[tight void]

friday night fr

[severe pond]

?

[tight void]

p awesome anyway

sorry for just asking

[placid oak]

bruh

[severe pond]

just dm them?

[tight void]

yeah my bad

but I have a question too

[placid oak]

shoot

[severe pond]

u shouldn’t directly ping someone

[tight void]

they don't have to answer the q

[severe pond]

ok still

[safe radish]

Please do not ping individual helpers unprompted.

[tight void]

yeah my bad

it looks symmetric

I used a markov chain

BUT

is there an easier way to prove it

and am I right

[eternal carbon]

the answer should just be 1/2 because encountering 55 vs 56 first should be equal

[tight void]

yeah

I got P_1 and P_2 as 0.5

cool ty

[eternal carbon]

that's a good sign

[tight void]

ty have a good night

and again Desync hope ur doing good

.close

[safe radish]

Channel closed

Closed by @tight void

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[severe pond]

rizz

[placid oak]

[safe radish]

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[sweet warren]

Can anyone explain how I got this wrong?

[light shoal]

what was your reasoning?

[sweet warren]

tbh idk

[light shoal]

so it was just a guess or what?

[sweet warren]

I just made an educated guess

[light shoal]

ok

well a hint to get started...

what's the sum of the four angles?

[sweet warren]

I was thinking to pick D or E

[sweet warren]

360

[light shoal]

yep

and what else can you say about the angles

[sweet warren]

A line is 180

[light shoal]

ok..

if you draw a picture of the situation.. some of the angles are equal to each other

[sweet warren]

yes

[light shoal]

now A and B aren't equal (B is 3.5 times A, that's given)

so A must equal C or D

[sweet warren]

yes

[light shoal]

doesn't really matter which, let's say A = C

[sweet warren]

ok

[light shoal]

in that case, B must equal D

do you agree?

[sweet warren]

Yup

[light shoal]

ok

so now you can simplify things

there are really only two angles

A and B

because C and D are the same as A and B

[sweet warren]

yes

[light shoal]

now the sum of the 4 angles is 360

[sweet warren]

ye

[light shoal]

can you write an equation involving A and B that captures this info?

[sweet warren]

let me try

2x = 2(3.5x) 🗿

[light shoal]

mmm, right idea

[sweet warren]

oh damn

[light shoal]

but there should be a 360 in there somewhere

[sweet warren]

oh

2x + 2(3.5x) = 360

there?

[light shoal]

yes

which you can simplify now

[sweet warren]

I can solve it now

[light shoal]

and solve for x

[sweet warren]

hold up

i got 40 for x

[light shoal]

yes

that's correct

[sweet warren]

damn

you made it so easy

Thank you bro

[light shoal]

yea it's all about setting up an equation

sure, cheers

[sweet warren]

wait

could I like plug in answer choices

with 3.5 x multiplying

[light shoal]

yea basically it would be similar to what you just did

you would take x and 3.5x as the two angles

and remember that there are two angles of each of those values

[sweet warren]

yeah

[light shoal]

and you would add up the results and see if they equal 360

but really it's probably faster just to solve the equation

[sweet warren]

ye

[light shoal]

also that plan might not work if they add a "none of the above" option

[sweet warren]

thanks bro

[light shoal]

sure

gl

[sweet warren]

.close

[safe radish]

Channel closed

Closed by @sweet warren

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[safe radish]

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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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[eternal adder]

Need help wit this

[eternal adder]

not quite understanding what to do here

[trim swan]

Do you know what they mean by local maximum/minimum?

@eternal adder

[eternal adder]

yeah it means as its written right like the local/in that area minimum and maximum

[trim swan]

sure, like a local max looks like the top of a hill, and a local min looks like the bottom of a valley

So, do you see any local maximums on this graph?

[vale oriole]

arent you usually given a domain for these questions?

[eternal adder]

theres two right?

[trim swan]

yes

[eternal adder]

the one at 4 and 2

[trim swan]

are you referring to the x-values?

[eternal adder]

wait i'm tweaking

[trim swan]

little bit lol

[eternal adder]

I was refering to the y values

mb

[trim swan]

ahh

yes

I'm tweakin too

The answer they're looking for is the entire ordered pair (x,y) though

[eternal adder]

\left(-4,\ 4\right),\ \left(6,\ 2\right) this no?

ok

what the flip copy and paste did not work

[trim swan]

yeah

[eternal adder]

(-4, 4) and (6,2)

[trim swan]

yep

[eternal adder]

cuz that didn't work

[trim swan]

Did you put (-4,4) in the first box and (6,2) in the second?

because the first box specifies the lesser x-value

[eternal adder]

yeah

oh shit

I dont need to put the parenthesees

[trim swan]

oh, it really rejected your answer for that? lol

fair enough, I guess

[eternal adder]

yeah

this program kinda sucks

it does that over little things

[trim swan]

yeah that's frustrating

but I suppose you can answer the local min as well then?

[eternal adder]

Yeah i got that

0, -2

thanks!

[trim swan]

👍 no problem

[eternal adder]

.close

[safe radish]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[kindred violet]

im doing a desmos project where i use various functions and relations to trace an image of my choice. im wondering how i could use a rational function to trace some part of this cat image

[kindred violet]

it can be any part

of the cat

[ruby bolt]

you could put equation of circles to trace the cat's eyes

[kindred violet]

oh does that count as a rational function?

[frozen marlin]

idk that counts as a function

maybe try the lower part of the face using sine/cosine?

or the mouth-thing

using like -sin^2 x

[kindred violet]

sorry i shouldve been specific, i have to do minimum 1 each of these 10 functions

and im specifically confused on the rational one

[kindred violet]

that helps tho tysm!

ik what to do for the trigonometric one now

[frozen marlin]

lol

NP

[kindred violet]

anyone know what i could do for the rational function?

[frozen marlin]

apparently a rational is a polynomial

divided by a polynomial

i have no idea tbh

[safe radish]

@kindred violet Has your question been resolved?

[kindred violet]

<@&286206848099549185>

[frozen marlin]

sorry

im juggling 3 channels at once

was waiting for u to come back

[kindred violet]

no worries at all !

[frozen marlin]

i think

u need to find a curve or smth

and then exploit that

ttbh i havent played around with rationals that much

okay how about this

draw the bottom of the face using a rational

nvm that doesn't seem to work

wait hold up im trying

[lean otter]

u could probably use the negative absolute within a certain domain to do the whiskers

that wld tick that off

[frozen marlin]

wait

about the whiskesr

that wouldnt be a function anymore

doesn't pass VLT

[kindred violet]

yesss i wrote that down earlier!

[kindred violet]

(highlighted ones are ones ive already done for clarification)

[lean otter]

ohhhh

[kindred violet]

ohhh shoott

i thought it woukd work

[frozen marlin]

yea

i think maybe ask ur teacher?

[kindred violet]

sure i can ask about that!

[lean otter]

u have to use trigonometric functions? why dont u make a circle w them

[kindred violet]

sorry let me send an updated list, ren helped me identify a trig function i can use earlier

[frozen marlin]

lol

ummm

wait log?

[kindred violet]

ratioanl is my biggest confusion,, maybe i should ask my teacher on monday lol if no one has ideas

yes

[frozen marlin]

okay

for exp or log

or rational

i think the tips of the ears

try functions like (x / x+5)^2

i think that might help

bc they all have sharp decreases at some points after some manipulation

[lean otter]

Yeah

I was thinking use exp/log on ears

it could work w some tweaking

[kindred violet]

wait i wanna verify

does this count as a rational function>

?

[frozen marlin]

DAMN

HOW DID YOU GET THAT CLOSE

[kindred violet]

it turned out reallt good

[frozen marlin]

yea i think so

[kindred violet]

I MESSED AROUND WITH THE ONE U GAVE

[frozen marlin]

LMAOOOOO

AHAHAHA

[kindred violet]

if thats rational thats literally perfect

[frozen marlin]

I DIDNT EVEN THINK THAT WOULD WORK LMAO

[kindred violet]

bc its just a polynomial over another right??

[frozen marlin]

what i was talking about

oh i just realized that looks cringe

okay

so if u set it to -(x^2/x+1) and zoom in

it becomes pretty much exactly like the cat's ear

wait no

yea sorry that's the correct function

@kindred violet how's it going rn

im tryna find one for log

[kindred violet]

trying this out rn!!

[frozen marlin]

okay

try $-0.5\left(\frac{x^{2}}{x+1}\right)$

[flat frigate]

ren

[frozen marlin]

i think that's a great approximation

[kindred violet]

ooo

should i set a domain restriction

[frozen marlin]

the bottom bit

[frozen marlin]

idt

wait no

yea set a domain restriction

do you see the bottom bit? the green one below the x-axis? that's what im talking about; just offset it a little bit

[kindred violet]

agh sorry how should i do that?

[frozen marlin]

umm

wait hold on

try this function

$-0.5\left(\frac{\left(x+2.8\right)^{2}}{x+3}\right)+3$

[flat frigate]

ren

[frozen marlin]

i think that should be around the points where the ears are

from there just tweak the values

i think -0.65 instead of -0.5 should do it

lmk if it works well

log is the hard one rn

[frozen marlin]

YES

but bad news

[kindred violet]

YESSS

[frozen marlin]

idt logs can be used for anything

except the ears.

[kindred violet]

oh no

[frozen marlin]

bc im experimenting with a LOT of graphs

i can't find anything remotely similar

[kindred violet]

maybe i can do one ear log one ear rational..??

[frozen marlin]

i'll try to figure smth out

[frozen marlin]

OH THAT'S SMART LMAO

[kindred violet]

if possible

[frozen marlin]

i think you can

[kindred violet]

OKOK

[frozen marlin]

LES GO

LISSA FTW

OMG

IM SO DUMB

I AM SO FREAKING DUMB

LOG GRAPHS

BECOME ESSENTIALLY FLAT

AT LATER INTERVALS

AND WE CAN OFFSET IT

SO WE CAN MAKE THE WHISKERS

USING THE LOG GRAPHS

can u send the image of the entire cat??

i need to figure out where to put the whiskers

@kindred violet

[kindred violet]

OHHHHH

YES

YES

[frozen marlin]

QUICK

SEND GRAPH

[frozen marlin]

okay

[kindred violet]

IS THIS FINE?

[frozen marlin]

yes

[kindred violet]

OK

[frozen marlin]

absolutely

tysm

$\ln\left(x+99\right)-5.4$ for the top left whisker

[flat frigate]

ren

[frozen marlin]

$\ln\left(99-x\right)-5.4$ for the top right one

[flat frigate]

ren

[frozen marlin]

.

@kindred violet try these!

@kindred violet are they good?

[frozen marlin]

YES

YES

LET'S FREAKING GO

NOW JUST PUT ON DOMAIN RESTRICTIONS

[kindred violet]

OKAYOKSY

[frozen marlin]

:D

next we need exp

imma work on those

i think 0.2x^2 - 2 or smth for the bottom of the face works

[kindred violet]

!!!!!!

PERFECT

[frozen marlin]

YES

LES GO

I LITERALLY CANNOT SEE THE LINES NOW LOL

[kindred violet]

ooh lemme try

[frozen marlin]

kk

[kindred violet]

PERF!!!

i can use a semicircle equation for the other half!! still need to add that

[frozen marlin]

NICE

[kindred violet]

lemme do that

rq

[frozen marlin]

:D

kk

try this for the mouth-thing: $-0.75\sin^{2}2x\left{\frac{\pi}{2}\ge x\ge-\frac{\pi}{2}\right}$

[flat frigate]

ren

[kindred violet]

added semicircle!

[kindred violet]

OKOKA

[frozen marlin]

gj

:D

this is going insanely well smh

how's it going

[kindred violet]

HMMMM

[frozen marlin]

??

that looks so bad

oofc

oof

i have an idea

use the semicircle bits for the mouth-thing

and try trig for the other half of the face

i think $-1.7\cos0.75x\ \left{0\ge x\ge-\frac{2}{3}\pi\right}$ works

[flat frigate]

ren

[kindred violet]

OKAYOKAYOKAY

[frozen marlin]

just downsize it

hide semicircle

and etc etc

??

is it good

enough

[kindred violet]

umm idk what im doing tbh

agh

[frozen marlin]

oof

wait

try this

i think this is good

$-3\cos0.4x\ \left{0\ge x\ge-1.25\pi\right}$

[flat frigate]

ren

[kindred violet]

OOH

OK

i’ll move semicircle to the mouth

or wait

which should i move to the mouth

[frozen marlin]

yes

[kindred violet]

semicircle or the exponential one we did

[frozen marlin]

semcircle to mouth

ummm

i think semi tbh

[kindred violet]

feel like im being silly but

i forgot how to make it shorter

[frozen marlin]

??

wdym

[kindred violet]

the semicircle

line

it’s too tall

[frozen marlin]

change -2.4 to smth else

;-;

and btw change the center

to be smth else

[frozen marlin]

noice

just change the center to be below 0

in the x

is it done?

[kindred violet]

so sorry can u explain this

[frozen marlin]

okay

[kindred violet]

think i’m having brain fog lol

[frozen marlin]

you have the semicircle equation

and it looks like the x value of the center is 0
what im saying is set it to smth like -0.5

[kindred violet]

OU

[frozen marlin]

nice

just duplicate it like thrice

[kindred violet]

ahh okay thank u

i think i’m done for the night my brain is fried

thank u so much for ur help!!!!

really you helped so much

[frozen marlin]

np

DM me if u need more help, this was fun :D

[kindred violet]

for sure thanks so much!!!! :D

.close

[safe radish]

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[river oriole]

In this problem by cauchy test should I apply limit directly on 2n^(1/n)

[drowsy karma]

the variable n should tend to infinity here, I know this task because I used it myself in our textbook

[drowsy karma]

then , to compute it, you will need to use stolz theorem

[safe radish]

@river oriole Has your question been resolved?

[river oriole]

Can I not use cauchy mean value here?@drowsy karma

[drowsy karma]

as I said, it is a sequence, there is a mistake in the notation, in the sense of what the variable approaches, , it is not a function, so you can 't use Cauchy Mean Value Theorem

[safe radish]

@river oriole Has your question been resolved?

[river oriole]

i use riemann integraion and got 4/e

[river oriole]

yes is this stolz theorem?

[river oriole]

i thought this riemann integration

[drowsy karma]

it is

stolz*

[drowsy karma]

riemann sum works here, too, but i always try to find more elementary method, if possible

[river oriole]

I got it now the whole process

thanks

[drowsy karma]

yvw:)

[river oriole]

.clsoe

.close

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[brittle zenith]

Hey

[brittle zenith]

Can someone help me

[languid stirrup]

sure

[brittle zenith]

thx

[languid stirrup]

What dint you understand

There's 3 sides givne

[brittle zenith]

no

[languid stirrup]

Lengths 9 4 5

Yah?

[brittle zenith]

its diffrent

[languid stirrup]

Wym

[brittle zenith]

i have to draw a triangle with only given (sin A = 0.6)

[languid stirrup]

uhm

[languid stirrup]

In this it says

ohh

[brittle zenith]

ye

i think 0.6=6/10=3/5

so sin is opp/hyp

[languid stirrup]

Wait

A triangle

Or just an angle?

[brittle zenith]

yes

[languid stirrup]

It says angle

[brittle zenith]

ow ye

[languid stirrup]

:/

[brittle zenith]

can you help me with that ?

[languid stirrup]

Have you learnt unit circle

[brittle zenith]

can you teach me ?

[thin bridge]

over complicating

[languid stirrup]

I don remember any other simple methods ngl

[thin bridge]

start with drawing a right triangle
label one of the acute angles as A

[brittle zenith]

oke

[languid stirrup]

And make the opposite to a and hyp 5 and 3?

[thin bridge]

i think 0.6=6/10=3/5
so sin is opp/hyp
and then label the sides in a way such that opp/hyp is 3/5 (or equivalent)

[brittle zenith]

just a sec i send a pic of what i have then

[thin bridge]

then pythag for the remaining side but probably isn't needed for what they're asking

[languid stirrup]

And the remaining angle

Or is that not needed

@thin bridge it specifically says 90

Sorry i mean

Angle

[languid stirrup]

Use pythag to find the other side

[brittle zenith]

oke

[languid stirrup]

And erase the side that's length is 3

[thin bridge]

make sure to mark that right angle appropriately

[languid stirrup]

Because it specifically says angle

[thin bridge]

Mezy pls stop

[languid stirrup]

Olay

Okay

[brittle zenith]

just a sec pls

[languid stirrup]

It doesn't say triangle tho 😦

[thin bridge]

Mezy pls stop

[languid stirrup]

Okay

[brittle zenith]

is this good ?

[thin bridge]

yes

[brittle zenith]

thx

[thin bridge]

`uh

actually slight mishap with your application of pythagoras

[brittle zenith]

ow

[thin bridge]

it should be 5^2 - 3^2

not 3^2 - 5^2

[brittle zenith]

my bad

i do got a other question if you dont mind

[thin bridge]

this seems to be just an excercise with a calculator

what's your issue with it

[brittle zenith]

ow its the wrong pic just a sec

I dont know how to

[thin bridge]

use inverse trig functions

[brittle zenith]

do i just have to do it like sin-1 (0,72134) ?

[thin bridge]

on your calculator they're commonly accessble with
2nd/Shift + "desired trig function"

[brittle zenith]

Is this good ?

[thin bridge]

minor mistakes with you wrote

[brittle zenith]

what did i do wrong ?

[thin bridge]

should be 3' not 33' for b)

[brittle zenith]

ow my bad, thx for youre help

.close

[safe radish]

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[safe radish]

@olive sorrel Has your question been resolved?

[olive sorrel]

.close

[safe radish]

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@cerulean birch Has your question been resolved?

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✅

[safe radish]

@cerulean birch Has your question been resolved?

[safe radish]

@cerulean birch Has your question been resolved?

[safe radish]

Channel closed

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[safe radish]

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[last spruce]

Do u have to integrate or just sub in 0 and 3 and find the difference

[last spruce]

<@&286206848099549185>

[limpid canopy]

I think you have to integrate taking the limit from 0-3

@last spruce

[last spruce]

Ok

.close

[safe radish]

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[lone arch]

Why do they integrate with bounds R_0 and R(T)?

[gentle herald]

Coz we know the value of R(T_0) is R_0

[lone arch]

So?

[gentle herald]

What shud it be in ur opinion

[lone arch]

Why not just the indefinite integral?

[dire fjord]

when you use indefinite integral

[hearty egret]

he have integrated in $[T_0,T]$ then he has used the u-sub with $R(t)=R$

[flat frigate]

everg

[dire fjord]

you will have a constant of integration

[hearty egret]

so the new domain becomes $[R(T_0),R(T)]$

[dire fjord]

using initial conditions you will get the same result

[flat frigate]

everg

[lone arch]

So we could use indefinite integration here too?

[dire fjord]

oh wait nvm i said something stupid

you dont have any initial condition that gives you the constant of integration

[lone arch]

when integrating, don't we have to have the same bounds on both sides?

[dire fjord]

so you are integrating both sides

but each side is being integrated wrt to a different variable

thus the bounds for each side are taken from the interval of each variable respectively

so R goes from R_0 to R and T goes from T_0 to T

[lone arch]

Why is the interval of R from R_0 to R(T)?

Why not from R_0 to 2* R(T) for example

[dire fjord]

why does T go from T_0 to T

[lone arch]

Hm, well it doesn't need to go from T_0 to T, does it?

[dire fjord]

if you want to change the upper bound T then all bounds will change too

[lone arch]

Thanks

.close

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[lone arch]

We can rewrite [\int_{R(T_0)}^{R(T)} \frac{\dd R}{R} = \int_{T_0}^{T} \alpha \dd T] as [\ln(R) \Bigg \vert_{R(T_0)}^{R(T)} = \alpha T \Bigg \vert_{T_0}^{T},] right?

[dire fjord]

yes

[lone arch]

Thanks

.close

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[dire fjord]

np

i didnt do anything

[safe radish]

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[sacred wadi]

hello

[sacred wadi]

why this is impossible ?

and

why this is possible ?

[desert pasture]

have you tried evaluvating both limits by hand?

[sacred wadi]

by hand ?

[desert pasture]

yes

[sacred wadi]

what do this meanµ

[desert pasture]

by working it out on a piece of paper manually

[sacred wadi]

yeah

[desert pasture]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[desert pasture]

here notice the limit is of the form 1^(infinity)

there's a theorm you can use here

[sacred wadi]

yeah e^ln(...

[desert pasture]

yup, if $f(x)^g(x)$ is of that form the limit is $e^ {g(x)(f(x)-1}$

[flat frigate]

Why am. I here

[sacred wadi]

yeah e^tanpi/2 a *ln(tanpi/4 )a)

[desert pasture]

why ln(tan(pi/4)x)?

[sacred wadi]

pi/2

[desert pasture]

no ln

[sacred wadi]

yeah x not in tan

[desert pasture]

no, what I mean is the limit is$e^{tan(\pi x/2)(tan(\pi x/4)-1)}$

[flat frigate]

Why am. I here

[sacred wadi]

where is the ln ?

[desert pasture]

evaluvate the power at x=1

[desert pasture]

there is no ln in the formula if I remember right

[sacred wadi]

there is

[desert pasture]

no, I just checked please show your sources

[sacred wadi]

we cant put base e if there is no ln

[desert pasture]

what? here's a derivation of the formula

[sacred wadi]

because (tan(pi/4)x)^tan(pi/2)x=e^ln(tan(pi/4)x)^tan(pix/2)

[halcyon marsh]

Why you making math so complicated 😂

[safe radish]

@sacred wadi Has your question been resolved?

[safe radish]

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[opal elbow]

hi

[opal elbow]

I'm a bit confused on how to find the upper sum and lower sum

this is the correct graph

but the only number on the Y axis is 2, so I really don't know what to put for the height of the boxes near the middle of the graph

i got eight for the upper sum via: f(2) * 2/3 + f(1) * 2/3 + f(2) * 2/3

[safe radish]

@opal elbow Has your question been resolved?

[safe radish]

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[rigid inlet]

you can use the fact that cos is 2pi-periodic to simplify that expression to cos(-pi/2)

well i didn't say that was the only step

you can then use the fact that cos is an even function to arrive at the desired result