help-27

[primal ferry]

lemme give it a look

sorry i went afk

@opaque talon are you still here?

[opaque talon]

yeye

[primal ferry]

rather than making a table let's go through this systematically

so the qs says we have a total of 750

450 females so ___

fill in the blank

[opaque talon]

300 males

[primal ferry]

i'll be giving you such blanks

perfect

now

of 750, we have 320 cat, 250 dog, and ___

[opaque talon]

ohhhh

180 others

[primal ferry]

perfect

now from these 180 we know ___

[opaque talon]

25 are males which implies 155 are female

[primal ferry]

perfect

going back to the females now. we now know that from the 450 females, ___ females own other

from the question we know 175 females have a cat, so we can deduce that ___ overall

and hence working out all other numbers

for the men

and just computing

[opaque talon]

155 females have others, 175 have cats, 120 have dogs

25 males have others, 130 have dogs 145 have cats

oh do i j use that info

do i not even need a diagram at all

YESSS

I got the first one right

thanks

made is seem so simple

.close

[devout snow]

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[primal ferry]

no problem mate

[devout snow]

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[gloomy moat]

Can anyone please help me with these?

[gloomy moat]

Prof wrote this after that: and said lambda1 = 1 and lambda2 = 3

This is the confusing part. What am I lloking at

[acoustic leaf]

those are the eigenspaces

which is the vector space of every eigenvector corresponding to that eigenvalue

(it is also the null space of A - lambda I)

[gloomy moat]

but how did he get those

no expl in the solutions is crazy ngl

[acoustic leaf]

once you have the eigenvalues you can just plug in lambda to A - lambda*I

and then find the null space of that matrix

that's the standard way for finding eigenvectors

[gloomy moat]

is there a video you know that could explain it to me

[acoustic leaf]

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

[gloomy moat]

thank you both

[acoustic leaf]

any video you find on finding eigenvectors should explain that sort of thing

[gloomy moat]

.solved i guess

[devout snow]

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[frosty cradle]

can someone explain how to do this i know the answer i just dont know how the guy has done it

this is the working out thing but i dont get it

[wicked turtle]

which step are you unclear on?

[frosty cradle]

first one

i assumed u had to cross multiply

so like

8(x+8)

and then 3(x+3)

but the layout is weird

[wicked turtle]

yea like line 3?

[frosty cradle]

kinda

[wicked turtle]

they're basically doing that, but in a more longwinded way

they're first putting the fractions on the LHS over a common denominator

then combining into one fraction

then multiplying both sides by the denominator

same result either way

[frosty cradle]

could you show me a simplified way to do it

if you dont mind

like the steps you would do

[wicked turtle]

i would just do like you said

multiply both sides by (x+3)(x+8)

on the left hand side you get 8(x+8) + 3(x+3)

and on the right side you get (x+3)(x+8)

[frosty cradle]

but what do you mean right hand side

[wicked turtle]

originally, the right hand side is 1

when you multiply the equation by (x+3)(x+8), the right hand side becomes (x+3)(x+8)

[frosty cradle]

so you wouldnt cross multiply first?

[wicked turtle]

that's essentially what i'm doing

tell me in words what cross multiplying means

[frosty cradle]

when i say cross multiply i mean like multiply the 8 by the x+8

and the 3 by the x+3

[wicked turtle]

well if you only do that, then you have changed the equation

like it's not equivalent to the original one anymore

because what you described is equivalent to multiplying only the left hand side by (x+3)(x+8)

but if you want the equation to remain equivalent to the original, you have to do that to both sides

[frosty cradle]

if i multiply it

like

lets say i multiply the denominators by 1

like for example

8 + 3 = (x+3)(x+8)

is that the same or no

[wicked turtle]

nope

because 8/(x+3) times (x+3)(x+8) is not just 8

it's 8(x+8)

[frosty cradle]

i think i understand it now thanks

really appreciate it

how do i close this?

[wicked turtle]

.close

[frosty cradle]

.close

[devout snow]

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[devout snow]

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[tribal zealot]

so i know this is probability but im just.. i watched the example video and it kind of left me with more quetsions than answers. at first i thought it would be like 11/29 but now i am not so sure

[stable wraith]

can you explain your reasoning?

[tribal zealot]

yes i can tryy :3

so its the probability that the thing its asking would happen over.. like

everything

so

maybe i added up the wrong amount of total people

im just overthinking it

[knotty sage]

woop

this works in a general case but we are talking about conditional probability

[tribal zealot]

in this example question this guy says that.. among the students with no cat, whats the chance they wuld have a dog? that would be 6/14. so is my question along those lines?

[knotty sage]

yep here the condition is given ( being not having a cat)

[tribal zealot]

and just forgetting the other category

in that casewait

wait

wrong side

or

so

[knotty sage]

nope still not correct

[tribal zealot]

ah

so could you explain further?

[knotty sage]

this would be the answer to
"if they got a cat what is the probability they have a dog"

[tribal zealot]

oh

i have to go now so this will probably time out but i am a little lost

ty for ur help, if i get a response while im gone ill check it 😞

[knotty sage]

just look at the row where they don't have the dogs

gives the probablity of having a cat

[devout snow]

@tribal zealot Has your question been resolved?

[tribal zealot]

thank you

[devout snow]

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[supple kite]

Answer key says b=3 but I got b=2.3811

[jade oak]

I think you put the anti derivative of b^2 as b^2x/3 instead of just b^2x

[supple kite]

oh yeah thanks

.close

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[sharp hound]

I can't figure it out
I just need the steps to do it
I know the equation (1 + f'(x)2)1/2

[sharp hound]

Looking for the length of the arc

[astral lodge]

you know $\int_a^b\sqrt{1+\qty(\frac{dy}{dx})^2}dx$ right

[woven radish]

ooaa

[astral lodge]

try to find dy/dx first

what do you get

[sudden yarrow]

oh hey it's you!

[sharp hound]

What's up bro

[astral lodge]

focus

lock in

[sudden yarrow]

did you got the answer for the star mass problem?

[sharp hound]

x(2/3)-1/4x(-2/3)

[sharp hound]

I think I did eventually

[sudden yarrow]

sorry for not helping, again

[astral lodge]

good

so now what are your bounds

[sharp hound]

You made an effort

[sharp hound]

1, 27

[astral lodge]

now just calculate

[sharp hound]

What do you mean?

[astral lodge]

the integral

[sharp hound]

don't I have to square it?

[astral lodge]

yeah i know

just do the process for the integral and tell me if you get stuck

[sharp hound]

squared is x(4/3) - 1/2 + 1/16x(-4/3)?

[astral lodge]

yeah

and this is looking hellish 🗣️

or wait, maybe not

yeah not too bad lol

[sharp hound]

plus one, gives + 1/2?

so integral of square root of that

[astral lodge]

yeah

[sharp hound]

x(2/3) + (1/2)^1/2 + 1/4x^(-2/3)

[astral lodge]

?

[sharp hound]

square root eof x ^ (4/3) is x ^ (2/3)?

[astral lodge]

we are on $\int_1^{27}\sqrt{x^{\frac43}+\frac12+\frac{1}{16}x^{-\frac43}}dx$

[woven radish]

ooaa

[sharp hound]

yes

[astral lodge]

this is just $\int_1^{27}x^{\frac23}+\frac14x^{-\frac23}dx$ from our previous form of the negated version

[woven radish]

ooaa

[sharp hound]

Why isn't the 1/2 there?

[astral lodge]

because of what we did earlier

[sharp hound]

?

add one, gives + 1/2, then square root of all of it

[sharp hound]

What happens to the 1/2?

[sharp hound]

@astral lodge

@sudden yarrow

[sudden yarrow]

hmm?

I'm not very good at calculus sorry

but, if you have the bounds, can't you just plug the numbers and solve it?

[sharp hound]

I have to integrate

I don't know how for this

well, I'll have to revisit it

.close

[devout snow]

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[sly hamlet]

wait what was the q

[astral lodge]

look at what we did earlier

[astral lodge]

this form

[devout snow]

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[restive river]

how did they get this answer?

[restive river]

does it become 2(root3)/3 = cosx and then inverse both sides?

x in this case because i set the thing equal to x

but in reality its theta

[sly hamlet]

this is arcsec

not recipricol

[vital sedge]

if y = sec(x)

then arcsec(y) = x

for what value is sec(x) = 2root(3)/3?

is the question

[restive river]

we didn't learn arcsec at all... are you sure i need to know it?

[sly hamlet]

-1 when with trig should never be interpreted as a recipricol

its another form of arc

[restive river]

yeah i guess, sec both sides..

[sly hamlet]

ehh dont think like that

[restive river]

ah so i can't assume sec inverse is just cos

[vital sedge]

sec(pi/6) = 3root(3)/2

[sly hamlet]

its not sec inverse is what im saying

[vital sedge]

so arcsec(3root(3)/2) = pi/6

[sly hamlet]

i assume you know arccos sin tan?

[restive river]

no.. never seen that before

we literally only learned the three basics sin cos and tan and their opposites

[vital sedge]

"their opposites?"

[sly hamlet]

youve been taught so weird wtf

[vital sedge]

like sec, csc and cot?

[restive river]

i mean cosecant secant and cotan

yes

[vital sedge]

weird they didn't tell u about the inverses

[sly hamlet]

idk why they taught you that before inverses

[restive river]

not sure lol, our teacher doesn't explain much at all just does easy examples and lets us do the rest

[restive river]

like sin inverse and stuff?

[sly hamlet]

because the recipriocls are useless

[vital sedge]

Yes

[restive river]

really lol?

[vital sedge]

barely ever used

[restive river]

so then what is the arc prefix before everything

[vital sedge]

yeah you know about sin^-1 and cos^-1?

[restive river]

like what does this arcsec even mean

[sly hamlet]

dont

[restive river]

yeah

[vital sedge]

arcsec = sec^-1

[sly hamlet]

think about arcsec

[restive river]

ohh

[vital sedge]

it's a better way to write it

[restive river]

so arctan is tan^-1

[sly hamlet]

think about 1/arccos

[vital sedge]

Yea

[sly hamlet]

yes, but remmber this is different from 1/tan

[vital sedge]

Yes

[restive river]

so 1/cos^-1?

[restive river]

mhm yeah

[vital sedge]

man I hate the ^-1 writing

[sly hamlet]

yes

[restive river]

what about it?

[sly hamlet]

have you learnt that

[vital sedge]

tan^-1 IS NOT 1/tan

[sly hamlet]

sin^-1 cos^-1 tan^-1

[restive river]

yeah

[sly hamlet]

arc is same thing

[vital sedge]

Ah great

[sly hamlet]

another way to write it is all

[vital sedge]

sec^-1 = 1/cos^-1

aka arcsec = 1/ arccos

[astral lodge]

ehehehe

NO

wrong

only sec(x)=1/cos(x)

[vital sedge]

@astral lodge

[restive river]

so circling back to this

[vital sedge]

oh wait maybe ur right

[restive river]

this is basicallly arcsec 2(root3)/3

[astral lodge]

yes

[vital sedge]

I never learned the idiotic notation and whatever sec csc and cot mean

[sly hamlet]

its the same thing

[vital sedge]

I learned it the right way

[restive river]

ok and how exactly does that help me solve this

[astral lodge]

so find a value on the unit circle s.t. secant will equal 2sqrt(3)/3

notice that $\arcsec(x)=\arccos(\frac1x)$

[woven radish]

ooaa

[astral lodge]

no 😭

[restive river]

so if i multiply both sides by sec

it gets rid of arcsec?

[astral lodge]

not multiply

[restive river]

or am i thinking about reciprocals again

[astral lodge]

just applying sec on both sides

[restive river]

so then in what way

oh ok

so 2(root3)/3 = secx

[sly hamlet]

oh yes my bad

[astral lodge]

yep

and therefore cos(x)=?

[restive river]

1/2(root3)/3?

because sec = 1/cos right?

[astral lodge]

eh

close

3/2sqrt(3)

[restive river]

wait why though

the 3 on the bottom fraction of the main fraction comes to the top?

ohh wait

nevermind i think i get it now

ok thanks everyone

.close

[devout snow]

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[devout snow]

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[brisk ocean]

How to do this, im quite unsure

[brisk ocean]

.close

[devout snow]

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[devout snow]

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[junior heron]

can someone tell me how to factor this with x method

[vast drift]

im not sure what x-method is but im assuming its some kind of substitution

note that what you have there can also be written as
$4(e^{4x})^2-4e^{4x}+1$

[woven radish]

Dootud

[junior heron]

yeah i just need to know how i can factor it

[vast drift]

does what i wrote look familiar ot you

[junior heron]

yes

[vast drift]

if you still find it hard to facotrise, try letting U = e^4x

[restive river]

it's to factor polynomials

[wintry sand]

yup

as a general / practical rule -->
if a complicated looking term, in this case e^4x, appears multiple times in the expression, susbtitute it

[junior heron]

yeah i totally get that

i have just literally forgot how to factor something when the leading factor has a coefficient that isnt 1

[wintry sand]

$$4u^2 - 4u + 1$$

[woven radish]

Klein Bottle

[wintry sand]

The method is called middle term split
there are three coefficients
--> 4
--> -4
--> 1

you have to find a combination of two integers in which the product is (4) (1) {basically the first and last number}
and the sum is -4

we look at the factors of 4:

1 and 4
2 and 2

but also

-1 and -4
-2 and -2

yup that's it! the very last one -2 * -2 = 4 and -2 + -2 = -4

therefore we write the middle term as:

[woven radish]

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

[wintry sand]

if you still didn't understand I recommend you go through this video:
https://www.youtube.com/watch?v=MtHEk6Yy6N4&t=2s

there is another alternative method attributed to Poh Shen Loh that you can try using

https://www.youtube.com/watch?v=QazK4JeEGVA

[devout snow]

@junior heron Has your question been resolved?

[devout snow]

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[stable sapphire]

If there was a 50/50 relationship where one partner gives 50% and the other matches 50% too as in efforts, finance, love, communication etc.

but what if it was 100/100 relationship where one partner gives 100% of their all and so does the other partner.

But technically isnt 50/50 and 100/100 but from what i explained it doesnt make sense for it to be the same?

[drifting mauve]

the ratio of whatever contribution they have made is 1:1 but the ratio of the contribution of what they have made (50%) and what they could make ideally (100%) is 1:2

i guess

maybe

the ratio is the same but their capacity to contribute is different

[pseudo basin]

why are you trying to apply math like this to relationships

[drifting mauve]

weird thing to ask indeed 😔

[stable sapphire]

ohh ok

[stable sapphire]

math can be more than just on paper. i come into this server asking unrelated math questions youd see on paper 😆 😆 😆 ty tho

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[restive river]

dont know where to start

dont have an iota

[lethal pollen]

Integrate to find the function for velocity and displacement

Uniform acceleration means constant acceleration

There's the start

[lethal pollen]

Acceleration is the rate of change of speed

Just because it has a lower acceleration doesn't mean that initially the speed wasn't higher

[restive river]

yh

[restive river]

i mean as in

omdz

i just automatically assumed that velocity was 0

[lethal pollen]

Yea

[restive river]

i give up

bro

how does this even work

like

i feel like it shouldnt

so we know

eventually

both Sa and Sb

are gonna be equal

so

car B

will have to make up

that 200 displacement

so let s be the same

so then why is it sA accounting the 200

[restive river]

shouldnt it be that sB has the extra 200

[queen orbit]

it does in the answer

oh

wait

[restive river]

it doesnt make sense

[lethal pollen]

t=0 refers to the time when B is 200m behind A

[restive river]

ok so?

[lethal pollen]

Nvm

[restive river]

wait

[lethal pollen]

Oh I see

The displacements are the displacements from their original positions

[restive river]

is that because A is 200 metres infront of B

but it doesnt make sense cos if they overtake B wouldve travelled more distance?

or something

[restive river]

yh

[lethal pollen]

So basically they're solving for when when the displacement of B from the original position is 200+displacement of A

Aka when B overtakes A

[restive river]

so

like u said

t=0

so then

their displacement

B=0

but A=0+200

[lethal pollen]

No

As I said the displacements are from their original positions

The reference points for A and B are different

[devout snow]

@restive river Has your question been resolved?

[restive river]

i dont get it

how am i wrong

like

what i said is the same no

no😭

[tough vigil]

pakistan or bangladesh?

[restive river]

pakistan

[lethal pollen]

Where oa and ob are the reference points

This is what oa and ob would look like

[devout snow]

@restive river Has your question been resolved?

[restive river]

yh thats what i mean

i understand

thanks

[devout snow]

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[winged saddle]

Is my laplace transform correct?

[winged saddle]

These are the questions for my subject Control systems.

<@&286206848099549185>

[devout snow]

@winged saddle Has your question been resolved?

[devout snow]

@winged saddle Has your question been resolved?

[winged saddle]

No

[devout snow]

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[rapid blaze]

from a point, a, due east of the foot of a tower, o, the angle of elevation is 40 degrees. from another point, b, due south of the tower and 100m from point a, the angle of elevation is 42. let the top of the tower be t, and the height of the tower be h metres, find an expression using triangle tob for the lentj of ob

[rapid blaze]

help please

this trig question is getting out of hand

[devout snow]

@rapid blaze Has your question been resolved?

[rapid blaze]

I tried to draw the the question out

but I got a problem

point T O and B are on the same line

which means to triangle is formed

no triangle is formed

there is 3 part for the question, I solved part a with my graph but it doesn't seem to work for part b

[winter patrol]

can you show what you drew

how are you getting T O B on the same line

[rapid blaze]

is there a camera on this

I'm using my computer to chat

[lime plaza]

Hm

[winter patrol]

you could reproduce your drawing on paint

[lime plaza]

Try to draw somewhere online and screenshot

[rapid blaze]

ok

[winter patrol]

have you made stuff like isometric drawings before?

representing 3d in 2d,
e.g. drawing cubes

[rapid blaze]

no

I'm in year 9

[winter patrol]

pretty sure you would've seen 2d representations of cubes before

[rapid blaze]

ye

how do you draw it then?

can you show me how to draw

[winter patrol]

note that vertically up isn't the same as north

[rapid blaze]

oh

how would you draw the diagram

could you do it on a paper and send an image?

[winter patrol]

consider this,
to make things easier to see the grey patch is the ground
you wouldn't need to include that in the diagram you draw
the north/south axis is drawn at an angle to distinguish it from vertically upwards

[rapid blaze]

oh

so you are suggesitng the diagram is 3d

[winter patrol]

you represent the 3d situation with a 2d drawing

[rapid blaze]

oh

is this better

[winter patrol]

ditch the vertical part below O

[rapid blaze]

ok

thank you so much

[winter patrol]

and draw in these to indicate your right angles

[rapid blaze]

ye I did that

[devout snow]

@rapid blaze Has your question been resolved?

[devout snow]

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[restive river]

I give up yo, I got k=4 please someone send working I took 5 pages and 5 attempts all wrong. Lovely

[ember schooner]

which one a b or c?

@restive river

[restive river]

C

[ember schooner]

ok first u have to calculate area of both triangles

so Bpa = 1/2 x 2 x root ak x 4a^2 x K / k^2 + 4a^2

and osp = 1/2 x 2root ak x k

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[frail beacon]

can someone tell me whats wrong in this method?

im getting non integral values for s5 after solving the recurrence relation

[devout snow]

@frail beacon Has your question been resolved?

[timber locust]

A(n) means that the 10 digit sequence is starting with 0/4?

If its so
it cannot be started with 0 ryt

[frail beacon]

it can

its not a number

just a sequence

[timber locust]

oh yea

[devout snow]

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[devout snow]

@frail beacon Has your question been resolved?

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as this action is irreversible, and this channel may abruptly lock.

[frail beacon]

.close

[devout snow]

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[fluid widget]

Arc LENGTH is 240.86 yd
One measure is 38 degrees
The circumference of the circle is 300 pi
What’s the second measure?

[pseudo basin]

wdym by "one measure"

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[fluid widget]

I don’t have a photo 😞 ill draw it out

https://cdn.discordapp.com/attachments/1331729189065326683/1364292126170550412/Untitled85_20250422122929.png?ex=6809238e&is=6807d20e&hm=ae211444cc8a24977dccdc2c99ed52c8b7dfb990b984d4351fd4399e57398bdb&

[pseudo basin]

ok

do you know how to relate arc length and angle

perhaps you have also learned about radians while you're at it

[fluid widget]

mhm

[pseudo basin]

ok right so

can you tell me the measure, in whatever unit (but you have to say which), of the 240.86 yard arc

[fluid widget]

Don’t we have to do
240.86 = x/360 x 300 pi?

I’m sorry I’m new to this 😭

[devout snow]

@fluid widget Has your question been resolved?

[devout snow]

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[manic sedge]

Can I take (a-1) common and then apply the properties of sigma?

Would that work

Or do I have to keep that a-1 inside the determinant

🥲

[supple knot]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/03%3A_Determinants/3.02%3A_Properties_of_Determinants

[manic sedge]

@supple knot I just want to know if this step is correct or not?

[supple knot]

every step you do should be justified by some property of summation or determinants

don't guess steps

[manic sedge]

Okay

I've just started studying determinants and the matrix is the next topic I can't understand clearly

[supple knot]

i recommend to start reading then

[manic sedge]

Okay

So I have to do it like that

[supple knot]

i have no idea what you're doing

[supple knot]

did you read this

[manic sedge]

I'm putting the sigma inside the determinant because of the sum property

.close

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[devout snow]

Please don't occupy multiple help channels.

[loud bluff]

Why is this wrong?

[rapid blaze]

how is the integrate form of 1/x+3 ln(x+3)

I think Le you forgot the +C

for the definite integral

indefinite

[loud bluff]

Wait is it not?

[rapid blaze]

I'm not sure

[loud bluff]

No I added the +C in the final answer

[rapid blaze]

ioh

[loud bluff]

Well the + K because they said to use K

[rapid blaze]

oh

[loud bluff]

It is

[rapid blaze]

huh

how

what

ye

[loud bluff]

<@&286206848099549185>

[lyric hornet]

you did your PFD wrong

[loud bluff]

What's pfd?

[lyric hornet]

partial fraction decomposition

[loud bluff]

Ohh but where? I double checked the calculations and everything

[lyric hornet]

well note the numerator is a degree 3 polynomial and bottom is degree 2, partial fractions doesn't even make sense if that is the case!! first you gotta do polynomial divsion

[loud bluff]

OHHHHHH

I just saw that

I see

[lyric hornet]

but other than that mistaken assumption the math is solid if it's any reconciliation lmao

[devout snow]

@loud bluff Has your question been resolved?

[devout snow]

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[supple knot]

It's just u sub if your f(x) is correct

Have you learned u substitution before

Did you read the hint

I mean the note

Show the entire work to this point

The purpose of the note is so you don't find the antiderivative of exp(-x^2)

[supple knot]

And probably integration by parts

[supple knot]

Yea try using IBP somewhere in the beginning

x^2 = x * x

[devout snow]

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[supple knot]

Dafuq

[devout snow]

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[lone stirrup]

Just wondering, in rational functions where you would like to find the asymptotes. In P(x)/Q(x). If P(x) has a higher or equal exponent number as Q(x), it has a slant asymptote, but if it's less it's a horizontal, correct?

[winter patrol]

not quite

[lyric hornet]

actually if (\text{deg}(P)>\text{deg}(Q)) then it's possible to get a quadratic asymptote! Consider (\frac{x^3+2x^2+3x+4}{x}) which has asymptotes (y=x^2+2x+3)

[woven radish]

PajamaMamaLlama

[winter patrol]

depending on the difference in degree, the asymptote may be non-linear,

equal degree, you'd also get a horizontal asymptote. (ratio of leading coefficients)

if deg(P(x)) is lower, that horizontal asymptote will be y=0

[lyric hornet]

as to why consider (\text{deg}(P)=n) and (\text{deg}(Q)=m) then (\frac{P(x)}{Q(x)}\sim\frac{x^n}{x^m}) if (n=m) then we expect the horizontal asymptote, if (n>m) then we can get slant or curvilinear asymptotes, and if (n<m) as ramonov said the aymptote will be (y=0)

[woven radish]

PajamaMamaLlama

[devout snow]

@lone stirrup Has your question been resolved?

[devout snow]

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[normal bolt]

More than one options may be correct

[ripe flax]

do u know how to do derivatives

especially trig functions

[normal bolt]

Yes but I am sticking in trig func

Stucking*

[ripe flax]

ok lets clean this expression up

we’re differentiating sqrt(sin(sqrtx))

which is just sin(x^(1/2))^(1/2)

correct?

[normal bolt]

Yes

[ripe flax]

do u know how the chain rule works

[normal bolt]

Yes

[ripe flax]

when you differentiate a function, you differentiate both the inside and outside

[normal bolt]

Yes

What do u exactly means by inside and outside

[ripe flax]

if something is in a function like sin(x), x is inside

[normal bolt]

Ok

[ripe flax]

an example of chain rule is the derivative of tan(3x)=3sec^2(3x)

we pulled the 3 out of the inside

now as u can see we have a behemoth of functions here

a chain rule inside a chain rule

[normal bolt]

Yes

[ripe flax]

lets substitute sin(sqrtx) for something like u

so we get u^(1/2)

we have the equation u=sin(sqrtx) and y=u^(1/2)

[normal bolt]

How y?

[ripe flax]

f(x) is basically y

its cleaner imo

i like leibnitz more than bernoulli

[normal bolt]

Yes

[normal bolt]

notations?

[ripe flax]

mhm

but now for the x and u we require another chain rule

so lets introduce another variable v

let v be the inner function, sqrtx

so v=sqrtx and u=sinv

clear?

[normal bolt]

Clear

[ripe flax]

now its easier to differentiate both in ur head

dv/dx=(1/2)x^(-1/2) and du/dv=cosv

[normal bolt]

Clear

[ripe flax]

multiply the derivatives out, the dv’s cancel snd u would be left with du/dx

now lets get to the equation with y and u

y=u^(1/2)

same thing, dy/du= (1/2)u^(-1/2)

now multiply the du/dx with the dy/dx so the du’s cancel

[normal bolt]

Where did you introduce this first

[ripe flax]

and dy/dx is basically just f’(x)

[ripe flax]

we let u be sin(sqrtx), so if y=sqrt(sinsqrtx) then y=u^(1/2)

then we let v be sqrtx

differentiate stuff, multiply and cancel the dv’s out, then cancel the du’s out then u would get dy/dx

do u understand now?

[normal bolt]

Clear till now

[ripe flax]

but hey

we multiply the dv/dx and du/dv and we get du/dx= cosv * (1/2)x^(-1/2)

what do we do we have another variable

not to worry, we previously stated that v=sqrtx so just sub sqrtx in

same for the variable u in dy/dx

[normal bolt]

Yes

[ripe flax]

and according to my calculations u should get b

[normal bolt]

Yes

B is answer but one more answer is there

[ripe flax]

ok now this is a matter of trig identities

do u know the basic ones

like reciprocal, pythagorean?

[normal bolt]

Yes ik basic ones

Yes

[ripe flax]

we know its not c cuz the cos just turned into cot for some reason

its not a cuz the sqrtx inside the sin is missing

its def d

but how do we check?

[normal bolt]

Yes it's f

D*

[ripe flax]

look at the denominator, answer d is missing a sqrtsinsqrtx

so would u agree we should divide the top and bottom of answer b by sqrtsinsqrtx

[normal bolt]

Yeah

[ripe flax]

ok now we fixed the denominator

but now we have a messy numerator

[normal bolt]

Yes

[ripe flax]

heres a fun trick

would u agree that cosx= sqrtcosx * sqrtxcosx

and its the same if x has a square root inside

so we have 1 sqrtcosx dividing by sqrtsinx but the other sqrtcosx just stays there chillin

[normal bolt]

Can u write this using latex

Anyway I understand

Let's not waste time

[ripe flax]

oh mb its sqrtcosx*sqrtcosx

so we divide the sqrtcosx by sqrtsinx

what do we get?

[normal bolt]

Sqrttanx

Cotx

sqrtcotx*

[ripe flax]

yes we get sqrtcotx

then the other sqrtcosx just stays there

so we have shown that d is also a valid answer

[normal bolt]

Clear

Thank you...

How much did you practice these stuff?

[ripe flax]

well i have maths lessons every day so every day

[normal bolt]

approx. how much problems?

[ripe flax]

idk, but we get thru 1 minor topic in 1.5 lessons

it depends honestly

[normal bolt]

Fine

[ripe flax]

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[normal bolt]

.close

[devout snow]

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[devout snow]

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[pseudo basin]

seems fine

[devout snow]

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[lime plaza]

lol

[devout snow]

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[tropic ferry]

can someone explain why these two numbers became... these...

[tropic ferry]

i am literally losing my mind right now lmao

[supple knot]

did you learn matrix multiplication?

[tropic ferry]

well i did... but i also kind didnt? might just have forgotten since im literally cramming all of IK in one night lmao

mind refreshing me on that?

[supple knot]

https://www.storyofmathematics.com/wp-content/uploads/2023/02/Matrix-Multiplication-Formulas.png

there are some simplicities you can do since you have rotation matrices and a lot of zeros

[tropic ferry]

im gonna try to calculate the thing from the screenshot myself and ill come back afterwards

thank you for pointing it out lmao

@supple knot is -sin710 in the calculator just the answer of sin710 * -1?

sorry if its a dumb question lmo

[supple knot]

-a = a * (-1) correct

[tropic ferry]

oh lmo cause i kinda thought it had something to do with shift+sin710 or something

thank you

yeah uh no i think its too time consuming to do this manually

thanks alot though!

[supple knot]

you really just need to matrix multiply the right column and bottom row. the top left 2x2 matrix is a product of rotation matrices so you just add up the angles

,calc 100 + 330 + 280

[woven radish]

Result:

710

[tropic ferry]

can i just somehow shortcut to using the angles with the x y coord?

iykwim

[supple knot]

no i don't

[tropic ferry]

instead of having to manually multiply A1 with A2 then A2 with A3

just add up all the angles then somehow get the final x and y ( the ones with decimals )

its just these two that im trying to get

[supple knot]

yea use the right columns formula here.

[tropic ferry]

but how

[supple knot]

wut

[tropic ferry]

like wouldnt it look like this if i added up the angles first

[supple knot]

the right column is just multiplication and addition

you use it twice since you have a product of 3 matrices

[tropic ferry]

multiply and add with?

[tropic ferry]

oh wait

[supple knot]

whatever a, l, b, o, c, r you choose

[tropic ferry]

i might just confused myself lmaoo

or actually i really still dont get it somehow

give me a sec

damn nothing

[supple knot]

to use this, you have to get H = product of two 3x3 matrices, but you have a product of three matrices

[tropic ferry]

so i do A1A2 first right?

then product of that with A3

[supple knot]

that's one way yes

[tropic ferry]

then the x and y ( c and f on the formula ) of A1 on my case should be 0 0 and with this formula, L and 0 should be 2 and 0 on the first part, but L and O only changes

oh wait

what i mean to say is

since g h p q are all zeros wouldnt it just go to 0?

oh man

i am so dumb

i imagined each pair inside the parenthesis were individual numbers

man

this is what it feels to live on coffee i guess

damn thank you so much @supple knot

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[hasty wave]

I’m trying to find the rel. extrrema, is this correct!

[supple knot]

correct but you could have found it sooner by simplifying first

x^2 - 1 can be factored

[pseudo basin]

hang on

is the denominator x - 1 or a badly written x - 7

i have suspicions

[supple knot]

oh 7 makes more sense

[heady current]

doesnt seem badly written to me

[pseudo basin]

a seven without its middle stroke is like an angel without wings

[hasty wave]

LMAO BYE

It’s 1 !

[heady current]

i have only seen like 2 ppl in my life write it with a middle stroke

[pseudo basin]

ok american

i suppose you'll now see a third

[heady current]

indian*

[pseudo basin]

i stand corrected

[heady current]

u do

[supple knot]

yea do this then

[pseudo basin]

i tend to associate sevens without a middle stroke with stereotypical american handwriting, hence the conclusion i jumped to

[heady current]

i hv only seen old teachers write it with a middle stroke

[heady current]

its alright

[hasty wave]

True I do the middle stroke too it looks cuter

[supple knot]

!topic

[devout snow]

Please read the channel description before posting, and stay on topic.

[patent nexus]

my teacher also force everyone write 7 this way, but also 1 with horizontal line in the bottom

[pseudo basin]

anyway yeah let's not turn this into handwriting debate

[hasty wave]

Should I factor first or take the derivative first? Like in general?

[pseudo basin]

so the original function is $\frac{x^2-1}{x-1}$ ?

[woven radish]

Ann

[pseudo basin]

in general, you should always simplify as much as you can before differentiation

[hasty wave]

ohhh I seee

My teacher always took the derivative first so I just followed his steps

[heady current]

ans should still be 1

right?

[hasty wave]

Yes it’s a 1

[pseudo basin]

yes, the function itself simplifies to just x+1 and so its derivative is 1

and so there are no extrema at all

[pseudo basin]

but also (x-1)^2/(x-1)^2 = 1 not 0.

[heady current]

thats what i was saying

[hasty wave]

Oh I didn’t mean 0 I meant 1 lmao

That’s my b

Ty everyone

not sure why every time I ask for help on here it’s either 7 ppl helping me or no one at all lmao

.close

[devout snow]

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[unique river]

Can anyone help me w/ this problem: An element with mass 670 grams decays by 21.8% per minute. How much of the element is remaining after 15 minutes, to the nearest 10th of a gram?
exponential decay I think?

[polar chasm]

Yeah, it's exponential decay.

do you know the formula for it?

Have you already tried anything?

@unique river

[unique river]

yeah Im just getting the wrong answer

im not sure of the formula

[polar chasm]

show your work

[unique river]

im on my pc w no camera i did it on paper i can try and see what i can do

[polar chasm]

Have you used this formula?

or anything similar?

What was the expression you got?

try typing it in text

[unique river]

hang on im timed working on another problem ill send in a sec

ok i think i got it thank you sm for the decay

I got my answer as 16.8

f(t)=670(0.7820)^15 = 16.75622173 = 16.8

the decay equation i had was written wrong

[devout snow]

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[amber sluice]

is drawing the best way to do this?

[supple knot]

yes

[amber sluice]

how would that process look like

im doing the one top to bottom

starting from the abs value function

[supple knot]

when x <= -1

plot |x + 3| - 6

when -1 < x <= 0

plot -4

and so on

[amber sluice]

i have the first two so far

[supple knot]

did you only plot the points

[supple knot]

this should be -6 <= x <= -1

the -6 comes from the lower limit in the integral

,w plot |x+3| - 6 for -6 < x < -1

[woven radish]

Results provided by WolframAlpha

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Requested by rie.mann

[amber sluice]

i

[supple knot]

follow this to restrict the domain https://www.desmos.com/calculator/voiheuzt6o

[amber sluice]

oh wow this is useful

[amber sluice]

can you help me understand the restrictions

wait i think i understand it

[supple knot]

.

you should also get another limit from the upper limit of the integral

from there it's just adding/subtracting areas of triangles/circles

[amber sluice]

u mean i break it up

in each point?

[stable flare]

I think if you just separate the integral by each piecewise function, that’s should work. So for example, you’d write an integral from [-6, 1] for |x + 3| - 6

[amber sluice]

.close

[devout snow]

Channel closed

Closed by @amber sluice

Use .reopen if this was a mistake.

[devout snow]

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.

[orchid umbra]

how do you ONLY maximize one part of the net/

[topaz crag]

Assume your first fold is x away from the border. What would be the equation of the cross section ?

[orchid umbra]

4-2x?

[topaz crag]

We’re talking about an area here

Can you draw what the cross section will look like ?

[orchid umbra]

lwokey idk what crosssection means

😭

[topaz crag]

Ok so if I slice a hotdog 🌭, what is the shape of the sliced área ?

[ebon coyote]

fok

I wanted to talk about Pringles cans

[ebon coyote]

You are close though - that's the length of the other side of the cross-section

With x being the width

[orchid umbra]

an oval?

[topaz crag]

Good !

[orchid umbra]

can you label what the cross section is on the diagram 😭

im so confused

[topaz crag]

Now if you slice the tunnel, what shape do you get ?

[orchid umbra]

i thought it was this

[orchid umbra]

half a cylinder?

[ebon coyote]

What's the 2D shape of the entrance of the tunnel

[topaz crag]

It’s gonna be a rectangle

[ebon coyote]

(tbf you didn't specify lol)

[topaz crag]

Do you see it ?

[orchid umbra]

oh

this tunnnel

ok

[topaz crag]

Ok

Now we want to maximise the area of that rectangle

Can you give a fórmula for that área ?

[topaz crag]

Use this

[orchid umbra]

wait

so

the cross section is just talking about the SA?

[ebon coyote]

The cross section is the 2D blue shape there (here a rectangle)

You're tasked with maximising the area of this rectangle

The diagram on the right in your question is trying to get you to visualise what 3D shape you're making (a kind of tunnel)

And you need to find the area of the "door" you would have to put there to seal the tunnel, essentially

[orchid umbra]

oh

would area be x (4-2x)?

[ebon coyote]

yeeee

[orchid umbra]

yay

and solve using that?

[ebon coyote]

Yeah, you want to find the value of x such that that area is as big as possible

[orchid umbra]

do we care about SA

[ebon coyote]

nope

...please do not take that out of context 🤣

[orchid umbra]

😭 🙏

whats the other equation tho?

[topaz crag]

https://tenor.com/view/miss-jay-jay-tyra-antm-americas-next-top-model-gif-4040226487718382132

[topaz crag]

There’s just one expression

[ebon coyote]

This is an expression

[ebon coyote]

expression*, mate

[orchid umbra]

oh

??

dont you uully need 2 for optimization

or is it bc

we already have everything in terms of x

[topaz crag]

https://tenor.com/view/bonjour-jimmy-falon-fireteam-french-gif-11356344199549968088

Im French we call It équation

[orchid umbra]

U french?

wow

[ebon coyote]

I thought it was a translation error yh

[topaz crag]

Ok now if f(x) = x(4-2x)

What is the domain of f ?

[orchid umbra]

um

[topaz crag]

Minimize f on that domain

[orchid umbra]

0

and

[ebon coyote]

Wait the difference exists in French as well

[orchid umbra]

x = 2?

honestly idk how to find the domain

[orchid umbra]

are you french too

[ebon coyote]

No, but I've studied it for some time

[topaz crag]

You did good

[orchid umbra]

are you canadian?

[ebon coyote]

I'm British

[orchid umbra]

oh

[topaz crag]

Now find the minimum

[orchid umbra]

ok

i got x = 1

[ebon coyote]

An expression is a combination of finitely many symbols (including variables and the like)
An equation expresses an equality between two or more such expressions

[orchid umbra]

bruh

wait

what

i only got the min?

theres 2 max?

no*

[ebon coyote]

Wait what?

[orchid umbra]

oh ok

[ebon coyote]

This is a quadratic, it only has a maximum turning point

[orchid umbra]

bruh what