help-36

[void valley]

Well, it's still applying n to dimensions and not areas

[sterile willow]

Also, thank you so much for taking your time with me, this is helping me a lot

What do you mean to the dimensions and not to the areas?

[void valley]

Firstly calculate cross-sectional areas (according to the given dimensions) and then use the modular ratio

and not the other way around

[sterile willow]

this?

[void valley]

Yeah, basically area of the rectangle

[sterile willow]

So like 12x100x2, 375x12

[void valley]

Yes, so this is for steel channel

[sterile willow]

and for wood it would be 375x88

[sterile willow]

so 351x12 for here too?

My bad, i thought 375 was the full width

[void valley]

No, because this for the steel channel

[sterile willow]

oh yeah true

[void valley]

Can you determine where the centers of gravity of these rectangles are? (You need to assume some origin for reference)

[sterile willow]

What do you mean?

[void valley]

You need y_i

i.e. the location of the center of gravity for each component

[sterile willow]

I think it's the middle of each

because it wasn't specified in the question

[void valley]

yes, generally for a rectangle the center of gravity is in the middle

always

[sterile willow]

So, now I have to find y1 through 4

[void valley]

This is how you divided them?

[sterile willow]

Yep

[void valley]

Great

Now, what will be y1?

Let's say the bottom edge is y = 0

[sterile willow]

Half of the width of y1? So 6mm?

[void valley]

Yes

y2?

[sterile willow]

175.5

[void valley]

No, you're reffering to x

Wait, for y1 it should be 50 mm

[sterile willow]

x?

Oh yeah!

My bad

[sterile willow]

then y2 would be 6

right?

[void valley]

Correct

[sterile willow]

y3 would be same as y1 which is 50

and y4 would be 44?

[void valley]

• 12

[sterile willow]

44+12?

or 88+12

[void valley]

44 + 12, because we have to include the bottom flange

[sterile willow]

is y2 for a2?

or a4?

[void valley]

y2 is 6 mm, and y4 is 56 mm

All in all:
y1 = y3 = 50 mm
y2 = 6 mm
y4 = 56 mm, right?

[sterile willow]

Ok fair, i'm sorry if I'm confusing you with my confusion

i think so yeah

[void valley]

Now, use this formula

Remember that for wood you are supposed to apply the modular ratio (for A4)

[sterile willow]

ok one sec, ill send the picture in a second

Like this? I added the red after cause idrk

[void valley]

Did you multiplied it by at the top and the bottom or...?

And it should be divided by 16.67 not multiplied, because A4' = A4/n

[sterile willow]

Now I'm confused again- why are we dividing?

[void valley]

We want an equivalent section made entirely of steel

it's as if we were transforming wood into steel

[sterile willow]

so I shouldnt have the red part there?

at all?

[void valley]

It's correct, but you should divide

[sterile willow]

This?

[void valley]

Good, last thing is that A4 is 375 * 88 = 33000

Because 375 is already the width of wood

[sterile willow]

ah- i think i did 351, mb

[void valley]

Perfect

You should get approximately 29 mm

[sterile willow]

one sec, lemme check

okai, so now I find out moment of inertia?

[void valley]

Yes

You have just calculated "c"

Well, almost

[sterile willow]

Quick question, if I say y bar instead of c, would it still be correct?

[void valley]

Yes, it's just notation

[sterile willow]

ah ok ty

[sterile willow]

because i need to take c away from something, right?

moment of inertia equation, right?

[void valley]

Yes, it's basically parallel axis theorem (Steiner)

[sterile willow]

I know that the numbers are wrong, but it's basically this, right?

I get AD, but the I section, I'm mildly confused

[void valley]

You mean for the bottom plate?

[sterile willow]

All of them

I forgot the equation for I

[void valley]

You know the formula for the moment of inertia of a rectangle?

[sterile willow]

Let me try to find it

[void valley]

Ah, I see

bh^3/12

[sterile willow]

OH

ok now that makes sense

Good lord, you..are saving me right now

But yeah, for the bottom plate, how do I go about doing that part?

[void valley]

e.g. for A1:
I1 = 12 * 100^3/12 + 1200 * (50 - 29)^2

for the bottom plate:
b = 375 mm (width)
h = 12 mm (height)
A2 = 4500
I = 375 * 12^3/12 * 4500 * (6 - 29)^2

[sterile willow]

a3 = a1

[void valley]

And for A4... we are supposed to apply the modular ratio

So basically, we can just plug A4' or however you denote it

And additionally divide I4 by n I guess

[sterile willow]

A4' = 33000/16.667
?

[void valley]

Yes

[sterile willow]

then Itr, I just add everything together?

[void valley]

So I4 = (375 * 88^3)/(12 * 16.67) + 33000/16.67 * (56-29)^2

Yes, Itr is the equivalent moment of inertia

For the transformed system

[sterile willow]

Then for the maximum stresses, I just insert all the letters in

Wait, before that, I have to actually find c

[void valley]

Yes, for steel you can use ȳ

I mean c = ȳ (for steel)

And for wood

The maximum stress will occur at the point furthest from the neutral axis

[sterile willow]

Of course substituting 29.04 for 33.15, they said 112, I'm guessing it's just 100?

[void valley]

Mhm, 100 - ȳ

That's it

[sterile willow]

Thank you SO MUCH

I spent a whole day searching it up online, I got nothing from anything, thank you so much <3

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[tiny gorge]

any thoughts?

[tall cove]

do I need to check every axiom 1-10?

[tiny gorge]

yep, generally you do, unless you're working with a subset of something that's already known to be a vector space

[tall cove]

okay yeah we learned about subsets today and I remember learning something like that

[tiny gorge]

in the case of the real numbers, for many of the axioms, you can probably just say "this holds because the property holds for real numbers"

e.g. you don't have to prove that addition of real numbers is associative

[tall cove]

yeah like this one axioms 1-3 I just did let u = 1, v = 2, w = 3

like what's the point in doing that is it not just obvious?

I'm honestly so new to the topic this is the first problem I tried solving

[tiny gorge]

for the real numbers, most/all of them will be fairly obvious and will result directly from the fact that R is a field

[tall cove]

hmm okay

do any of these seem more difficult? perhaps I can try one that's harder and you check my work?

[tiny gorge]

do you have any thoughts on 8?

[tall cove]

hmm

well standard addition and multiplication so that's easy enough

maybe the multiplication stuff will end up weird?

[tiny gorge]

<@&268886789983436800>

[tiny gorge]

check whether it's closed under addition and scalar multiplication

[tall cove]

"closed" gets mentioned a lot, what does it mean?

[tiny gorge]

axioms 1 and 6

[tall cove]

as in equal?

[desert mantle]

no...

[tiny gorge]

axiom 1 can be paraphrased as "V is closed under addition"

just means that the sum of two members of V is in V

[desert mantle]

"closed" generally means that you do some thing (i.e. here, add two elements from the set) and the result is again in the set

[tall cove]

okay so how would I start? for like 1 and 6 for example

let u = (u1, u2) v=(v1,v2)?

[tiny gorge]

try reasoning without writing stuff down

(until you want to write a formal answer)

is the sum of two invertible matrices invertible?

[tall cove]

yes

[tiny gorge]

how about the product of a scalar with an invertible matrix?

[tall cove]

that is as well

[tiny gorge]

sadly both are false

[tall cove]

LOL no way

[tiny gorge]

what if the scalar is zero, for example

[tall cove]

oh, right

[tiny gorge]

and what if you add A and -A

[tall cove]

oh...

ouch

[tiny gorge]

haha that's why i suggested this one, it's not as obvious as the real numbers example

[tall cove]

okay so 1 and 6 don't work so is there even any reason to check the other ones?

[tiny gorge]

oh another quick thing to check, always

a vector space has to contain the additive identity, which would be the zero matrix for matrix addition

but the zero matrix is not invertible

[tall cove]

jeez, these are all words I've heard in lecture but they just didn't click

[tiny gorge]

nw, they'll click a lot more after you do more examples

[tall cove]

do you suggest I try to do more problems or like watch videos or read up on the topic?

[tiny gorge]

try problems, i think it's the best way to learn

i know it's tedious going through ten axioms but it's useful

(at least while you're first learning)

[tall cove]

okay for sure

I'll make note of everything you said and try some more problems, thanks for the help

[tiny gorge]

sure, feel free to ask here if you want to check your answers or if you get stuck

[tall cove]

will do

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[wispy shore]

Is there any easy way to solve this???

[sly forum]

trig ratios?

[small urchin]

easiest way is to use trigo

[wispy shore]

Yeah but idk how to solve jt

[small urchin]

okay so do u know any trigo identities?

[wispy shore]

Uhh no??

[small urchin]

okay so when we refer to basic trigonometry, its often explained in terms of right angled triangles

so when u have a right angle triangle with one of the non 90degree angles as theta

u can use that to determine other angles and sides

there are three fundamental (actually six but these r the basic three) identites

[wispy shore]

Ah sin cos tan?

[small urchin]

yes

[flint hemlock]

Use change of subject with sin cos tan

[small urchin]

u can just use sinθ = opposite/hypotenuse

[final saddle]

@wispy shore Has your question been resolved?

[umbral grove]

There is an easy way

let me do it

[wispy shore]

Ok

[umbral grove]

you can use calculator?

[wispy shore]

Yeah the t1 84

[umbral grove]

alr

[small urchin]

bruh idt u need a calci

[umbral grove]

i wish you can understand it

[umbral grove]

its just a question, i dont need it

[small urchin]

u overcomplicated it bro

its just sinθ= opp/hyp

so sin30=5/c

c= 5/sin30

so c =10

[umbral grove]

Speaking of which, just use the answers from the parts in the Pythagorean theorem

[small urchin]

then tan30=opposite/adjacent

so tan30=5/b

b=5/tan30 = 5root3

[umbral grove]

yeah, that's correct but not the only way, maybe he doesnt know bout tan or idk

i only use sin, i think its easier... for me

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[tranquil pine]

I dont even understand the question, what shud i do for i and ii?

for i, I sub x =10?

[spark swan]

for i, they want you to find when the investement will increase, so when the function is increasing

specifically in the first 10 years

[tranquil pine]

oh

[winged mesa]

is this 12board sample paper

[tranquil pine]

no

[spark swan]

can you solve the problem now?

[tranquil pine]

ye

what abt ii?

its increasing in (-inf,3) and (5,inf)

[spark swan]

i think lean patch would refer to a period of poor performance

so when the investement is decreasing

[spark swan]

okay so where is it increasing between 0 and 10

[tranquil pine]

then its (3,5)

[tranquil pine]

(0,3) and (5,10)

ok cool, thanks for helping

.close

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[uncut topaz]

What part are you stuck?

[flint estuary]

Hmmm

[south raft]

I am not technically a helper but i have some ideas on how to solve this

[flint estuary]

This I think this type came in our school last year

[south raft]

Should I give an idea?

Alright I'll write it here so it may take a bit of time

[small grail]

!status

[final saddle]

What step are you on?
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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

[small grail]

anyone helps is helper

[final saddle]

@neat wolf Has your question been resolved?

[south raft]

I have simplified it a bit but I may need a fresh perspective to go on

Please note that it all equals zero at the end

If anyone has difficulty understanding what I wrote, please ping me

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[ivory lynx]

Hi there, when using rules of logical equivalence to derive if its a tautology, contradiction, or contingency, how do we know if its a contingency?

[ivory lynx]

cause for tautology and contradiction, you just need to see if the truth value after deriving is either T or F

[proper dagger]

isn't a contingency just a statement that is neither a tautology nor a contradiction?

<@&268886789983436800>

[ivory lynx]

yeah

but we're using rules of logical equivalence (de morgans, associative etc etc) here instead of using truth tables

[proper dagger]

oh sorry, missed that part

[ivory lynx]

the truth tables should be proof that its either a contingency, tautology and contradiction in my worksheet

i acc asked this because im confused

on this part

((p v q) ^ r) <-> (p v (q ^ r))

heres my solution for now

truth table says its a contingency

[desert mantle]

are you asking how we can easily read off from some formula after doing some rules of equivalence whether the formula is a contingency or not?

if yes then this is just strictly not possible cause its basically the SAT problem

which is famously very hard

otherwise I'm not sure what you mean by your question

[ivory lynx]

i guess its this one

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[final saddle]

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[static oak]

Can somebody check this for me specifically 10 a

[dusty quarry]

whoa there. so 3sin(theta) = -3, which gives sin(theta) = -1. where did the sqrt(2) come from?

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[dusty quarry]

?

[static oak]

Special right triangles

[dusty quarry]

what made you think I'm here to ask a question lol

[void valley]

Oh, mb

I'm sorry

[dusty quarry]

that does not explain what gave you the right to stick a sqrt(2) to the value of sin(theta)

[gritty chasm]

??

[dusty quarry]

that changes the question

[static oak]

Because six is y/r

[smoky bramble]

should we revisit the unit circle definition of sin and cos

it might be necessary here

[static oak]

I don’t know how else I would solve the problem because I would not know the radius or any other side lengths

[smoky bramble]

what's y and r supposed to mean?

[static oak]

Uhh I think opposite and hyp I can’t spell that full word

[smoky bramble]

have you gotten a simplified expression for sin(x)?

[static oak]

I think

[smoky bramble]

what did you get

[static oak]

Sin(O) = -1

In my teachers answer key for a different homework but same problems we use the special right triangles

[smoky bramble]

mhm

and which angles satisfy that equation?

uh do you know the unit circle definition of sin and cos

[dusty quarry]

,rccw

[smoky bramble]

huh

[static oak]

The angle values? I think

[smoky bramble]

looks like you don't understand the unit circle definition completely

[static oak]

I mean this is how we did it with other similar problems in clsss

[smoky bramble]

say, there is a circle of radius 1

[smoky bramble]

its not a special triangle this time

plus, do you want to rely on only one tool to do the job?

[static oak]

how would we know? Like how would I know a test

[static oak]

My teacher said not to use the unit circle right now

Like we haven’t done it at all this year

[smoky bramble]

don't listen to them

teaching trigonometry without the unit circle is the same as teaching how to run without teaching how to walk

[static oak]

I mean I kinda have to for the test and this is the way I know

[smoky bramble]

ok continuing from here

this circle has a center, of course yes

now, if we chose a random point on the circle

say, 45 degrees, and draw a line from the center to that point

we then get coordinates

[static oak]

I just don’t know how to do that I did it some last year and I can’t really learn it now because my test is in a little over a hour

And I don’t have the unit circle memeorized

[smoky bramble]

.... THERE IS ZERO MEMORIZATION INVOLVED

[smoky bramble]

say (x,y(

then x=cos(theta)
y=sin(theta)

[static oak]

I just don’t know if I can learn a new method 😭 and be confident with it

[smoky bramble]

see this diagram?

there exists a x and y coordinate for the random point selected

[static oak]

Yea

Oh yea I know that part with x and y we just labeled side lengths with x and y

[smoky bramble]

say, there is theta2

theta 2 is labeled a bit badly but you get the picture

can you determine the sign (positive or negative) of the (x,y) coordinates?

[dusty quarry]

what a bad take

[static oak]

I mean I find it easier tbh

I also know she adjusted the unit slightly because none of us really learned square square roots well. Like we know the basics but nothing beyond that because of Covid. Idk if that’s related

[static oak]

Not without any side lengths or angels

[smoky bramble]

say theta2 is between 180 and 270 degrees

what would be the sign of (x,y)

[static oak]

I don’t know

[smoky bramble]

answer: -x, -y

[static oak]

That would be quarter 3 so x and y would be negative

[smoky bramble]

the answer is what you see

exactly what you see

[smoky bramble]

going back to this

[static oak]

It would be defintly in quarter 4 at least

[smoky bramble]

in a unit circle of radius 1, if the center is at (0,0) what can you infer?

[static oak]

And maybe quarter 3

[smoky bramble]

cos = x values of an angle
sin = y values of an angle

[static oak]

Because q3 and 4 is when y is negative

[smoky bramble]

consider watching youtube videos on this topic

[static oak]

Wait why am I wrong

[smoky bramble]

eh i worded the question a bit badly

the answer is that it can be in quadrant 3 only

assuming 180 and 270 aren't counted as 2nd and 4th quadrant

its not enough time to introduce a new topic sadly

[static oak]

I’m just trying to figure out which special right triangle it is because I realized the 45 45 doesn’t work

Wait

[smoky bramble]

the special triangle is a 0 degree triangle

in this case

[static oak]

Is this correct

[smoky bramble]

correct

[static oak]

Yay

[smoky bramble]

the minus solution is redundant tho

it's fine to keep it ig

[static oak]

Our limit is this

[smoky bramble]

its kay then

[static oak]

Limit is not the right word for it

I think

[smoky bramble]

domain is the word

[static oak]

yea

[final saddle]

@static oak Has your question been resolved?

[dusty quarry]

hey, might wanna close this before you open a new one @static oak

[static oak]

Whoops thought it was closed and clicked x instead of the checkmark I’ll just send it here

I’m not sure why I got this wrong. I know sin starts at the midline at the start of the period then moves up unless it is reflected over the x axis. But my teacher started with going down at the start of the period

[dusty quarry]

<@&268886789983436800>

[final saddle]

@static oak Has your question been resolved?

[final saddle]

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[heady sedge]

can you help me graph this function with absolute value

[final saddle]

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.
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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

[heady sedge]

1

[crude idol]

so you know how to graph x^2-2x right?

its (x-1)^2-1

[heady sedge]

not really

im bad at maths clearly

[crude idol]

so do you know how to graph x^2

[heady sedge]

yes like a U

[crude idol]

yes yes correct

so (x-1)^2 is translating the graph 1 unit to the right

the x^2 graph i mean

[heady sedge]

okay

[crude idol]

and (x-1)^2-1 is translating this graph down one unit

[heady sedge]

yeah okay

but can you explain one more time why

x-1

[crude idol]

think of it like every x value is pushed back by one so they need to compensate by going one more unit to the right

[crude idol]

so that means everything moves to the right

[heady sedge]

ohh okayy

[crude idol]

so you know how to graph x^2-2x now

[heady sedge]

yes

[crude idol]

ok so

you know that absolute value makes every value positive right?

[heady sedge]

mhm

[crude idol]

so it will flip every negative to the positive

the positive stay unchanged ofcourse

so the part that is negative is reflected through the x-axis

[heady sedge]

so with the absolute value it moved

to the positive

[crude idol]

ye

the negative part is now positive so its flipped

[heady sedge]

yep yep

[crude idol]

so -|x^2-2x| now turns all of our value negative

[heady sedge]

so it flips it

[crude idol]

yes

so after that adding three means the graph goes up by 3 units

yes

this is the final result

thats basically it

[heady sedge]

so thats it i dont need no formula or anything

[crude idol]

i mean you only need to graph it

then yes this is how you graph

[heady sedge]

okay bet

[crude idol]

this is graph transformation, so yes, no formula

[heady sedge]

thank you man

[crude idol]

alr np

[small grail]

!done btw

[final saddle]

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@heady sedge Has your question been resolved?

[heady sedge]

oh yeah sorry

!done

[final saddle]

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[heady sedge]

.close

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[candid pulsar]

huh??

[candid pulsar]

clearly |A|max = 27
how is it so clear here

[loud sundial]

hadamard's inequality

[https://en.wikipedia.org/wiki/Hadamard's_inequality]

or just vibe it cause jee lol

[candid pulsar]

once again diving into actual matrix knowledge thats conveniently gatekept from me

[small grail]

avg jee shit

[candid pulsar]

either some forbidden knowledge or whatever this is

[candid pulsar]

most reasonable direction i can try for this question is to assume the rows to be vectors lying on x + y + z = 3 and try maximizing the volume bound by the vectors

and in this case somehow volume is maximized when the vectors are orthogonal to one another

im not even going to

.close

[final saddle]

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[candid pulsar]

thanks for the help though

[final saddle]

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[oblique estuary]

the greyed out parts are the ones I currently have selected

I believe those 2 are correct

I think the second and third down from the top are wrong

but it's not a combination of the second/third from the bottom with the top and bottom ones

so I must be doing something wrong

[small grail]

right

[small grail]

5th is off

[oblique estuary]

wait really

oh

[small grail]

its indefinite integration

[oblique estuary]

5th is second from bottom

[oblique estuary]

right

[small grail]

oh i mean 6th

[oblique estuary]

Ohhh it doesn't have +C

I see

u right

cool

got it right

[small grail]

[small grail]

2nd is intimidating

[glossy zephyr]

4th is to be argued tbh.

But id personally say its incorrect

[small grail]

hm.. they did not define v

[glossy zephyr]

doesnt really matter

[oblique estuary]

it marked it as correct

idk

[glossy zephyr]

ok ig

[oblique estuary]

I'm just tired of calc 2 and I'm only a week in 😔

[small grail]

why

[glossy zephyr]

the problem is that as a function $\sin: \R \to [-1,1]$

[glossy zephyr]

and defining arcsin as its inverse we have
$\arcsin : [-1,1] \to \R$

[glossy zephyr]

So inputs that surpass 1 dont make much sense.

but anyways, numerically its logical

[oblique estuary]

idk I just want to actually get this homework done because I'm trying to start the year off by not just checking the TeX code for answers

because that works and it means I don't spend 6 hours on a homework assignment but the class is graded 90% quizzes and exams

so I gotta know the stuff

[small grail]

right

[glossy zephyr]

also, 5th was correct too right?

[oblique estuary]

yep

[glossy zephyr]

remember chain rule

[small grail]

what flaw do you see

[oblique estuary]

ah

the one I didn't memorize very well

[glossy zephyr]

no, 5th was plainly right

[glossy zephyr]

tbh if you dont recall chain rule youre gonna be struggling a lot.

[oblique estuary]

ikr

[glossy zephyr]

its arguably the most important

[oblique estuary]

yeah

[tender dust]

please kill me

[oblique estuary]

why

that was like

one of the easier questions

[tender dust]

bro im 9th grade tf is this

[oblique estuary]

i thought

[oblique estuary]

ah

go to #❓how-to-get-help

if you're looking for help

[tender dust]

noono but what is this "l" smth

yk what

nvm

haev a nice day

[oblique estuary]

you'll learn in calculus one

[glossy zephyr]

they are called integrals / antiderivatives

[oblique estuary]

you too

[small grail]

you will die one day

[tender dust]

thanks for reminding e

[oblique estuary]

realistically soon too, you have less than 30,000 days left

(probably)

anyways

[marsh mountain]

a little morbid

😭

[oblique estuary]

ohhh the chain rule is easy

I just needed a refresher

been a bit over a month since calc 1

so am trying to get back into math mind

[oblique estuary]

:)

life's short

do math I guess

aight, so, chain rule

dx/dy = dx/dt times dt/dy

plus the other parts of chain rule

but idk what exactly that does here

y'know

[small grail]

eh i just do subsitution then again

[oblique estuary]

I guess I could just go to to office hours

[oblique estuary]

OH
yeah

Idk why i didn't even think of that

that was stupid of me

substitution

is like the first thing to try

[small grail]

?

[empty herald]

ln(x)=t

[oblique estuary]

I'm used to using u

but yeah

alr alr

lemme cook (then get confused) and will report back when I get it wrong

alr so integration of ln(u)/u times du

[empty herald]

i believe integration is the right word lol

[oblique estuary]

oh

prolly

[empty herald]

summation is like E

[oblique estuary]

brain is fried currently, this is hour 5 on this assignment

[oblique estuary]

yep

you right

it's the diagonal lookin E

anyways

hm

I can integate ln(u)

[small grail]

m side way$\sum$

[soft zealot]

scoob

[oblique estuary]

I just don't exactly know what to do with the u in the denominator

[oblique estuary]

right

diagonal lookin E

[empty herald]

now substitute u as something

lnu

mb

[oblique estuary]

what

[oblique estuary]

why

we haven't ever done that before

[empty herald]

ln(u)=p

[small grail]

ye substitute thesubstitution

[oblique estuary]

wha

[empty herald]

then dp= 1/u*du

[oblique estuary]

right

okay

[small grail]

layered substitution

[oblique estuary]

so then you get p

[oblique estuary]

yeah

makes sense

[small grail]

i hope this violates nothing tho

[oblique estuary]

I just wish that professors taught stuff before like

putting it on homework

[oblique estuary]

what do u mean

[small grail]

nothing

[oblique estuary]

We just have no money for student math tutors because of budget cuts so

:P

here I am

[marsh mountain]

it doesnt

[oblique estuary]

fuckin Ohio

[marsh mountain]

since we’re not defining bounds anyways

[oblique estuary]

oh math wise

okay so

[marsh mountain]

there is also just one substitution available

rather than two of them

[oblique estuary]

is the integration of p times dp just p

or is it p^2

I'm pretty sure it's just p

[marsh mountain]

well

[oblique estuary]

but just wanna check

[marsh mountain]

whats the integral of xdx

[oblique estuary]

dang is it neither

x

right

[marsh mountain]

er

[oblique estuary]

damn

[marsh mountain]

no

[oblique estuary]

I'm cooked

[empty herald]

@oblique estuary what are you currently studying ?

[marsh mountain]

i wanted to see first if it was the letter tripping you up lol

[oblique estuary]

ohhh x_C

[marsh mountain]

anyways

[oblique estuary]

+C

[marsh mountain]

still no

[oblique estuary]

first week of Calc 2

[empty herald]

is that in cllg ?

[oblique estuary]

this is supposed to be a review homework

[oblique estuary]

yes

[marsh mountain]

so yk how antidifferentiation (integration ish) is the opposite of derivatives right?

[oblique estuary]

right

[marsh mountain]

lets rev for a sec

[oblique estuary]

we did that

[marsh mountain]

whats the derivative of x

[oblique estuary]

1

[marsh mountain]

okay sick

[oblique estuary]

Chat i'm slow but I know that

I just forgor other basic stuff

[marsh mountain]

when we integrate 1dx what we get

[oblique estuary]

💀

[oblique estuary]

x

[marsh mountain]

ur chill np

[marsh mountain]

sick

so when we integrate xdx

we’re asking

what do we differentiate to get x

[oblique estuary]

.5x^2

[empty herald]

x^2*

[marsh mountain]

peak

[empty herald]

what we get on differentiating x^2 not x

[marsh mountain]

i said what do we differentiate to get x

[empty herald]

wait mb i forgot to read to

[oblique estuary]

real

[marsh mountain]

LOL

[oblique estuary]

illiterate gang

[small grail]

chill guys

[marsh mountain]

illiteracy type shi

https://tenor.com/view/jonah-hill-i-guess-bro-bro-i-guess-ok-man-gif-6278747235421617446

anyways

you basically got it

cassie

[empty herald]

shttt i might have Dyslexia

[oblique estuary]

oh so it's just .5p^2

[hidden ore]

yall good here...

[marsh mountain]

ye

[oblique estuary]

I think

[marsh mountain]

then just undo the substitutions one by one

[hidden ore]

lmao

[oblique estuary]

yeah

blanket is good

[hidden ore]

she thinks...

[oblique estuary]

one braincell

[oblique estuary]

right

okay

[marsh mountain]

we chill here

[hidden ore]

can i see the problem?

[oblique estuary]

lemme cook

nah we got this

I'm sure other ppl need help

[small grail]

https://tenor.com/view/cook-breaking-bad-hoodie-gif-13536614

[marsh mountain]

$$\int\frac{\ln\ln x}{x\ln x} \dd x$$

[small grail]

yo

d

[marsh mountain]

oops

[soft zealot]

blanketism

[marsh mountain]

yea

im typing on my phone lmao

[empty herald]

yo tf you mean illiteracy

[small grail]

aight

[hidden ore]

blanket

how do u use the teXit bot.

[marsh mountain]

yo

[small grail]

not literate

[oblique estuary]

dangit i thought the answer was

.5 (ln(ln(x)))^2

I was wrong

[empty herald]

since when is knowing english equivalent to litereacy

[hidden ore]

@marsh mountain

[small grail]

when englanders colonized most people

[oblique estuary]

ANYWAYS

[marsh mountain]

did they perhaps want fractions?

that should be the answer

[oblique estuary]

no

[marsh mountain]

you can use the texit bot by writing latex then enclosing it in inline/display delims

[small grail]

blanke q is lost pin it

[oblique estuary]

+C

whoopsies

:3

[marsh mountain]

ah i assumed you already put it lol

nice!

[oblique estuary]

alrighty

so I should be able to just

[marsh mountain]

fwiw, you remember how you were asking about chain rule?

[empty herald]

knowing a language isnt equivalent to litereacy , most chinese cant speak english so based off ur def they are illeterate ?

[marsh mountain]

you couldve done this problem with one substitution, u = ln(ln x)

[empty herald]

brother there is lnx below too

[marsh mountain]

differentiate it?

[oblique estuary]

got the second part

[empty herald]

yea mb that works

[oblique estuary]

ahhh

[marsh mountain]

anyways

[oblique estuary]

that's what I must have been meant to do

[marsh mountain]

yeah

[oblique estuary]

instead of layered substituition

[marsh mountain]

its fine though

review

[oblique estuary]

yep

I mean layered substition isn't review for me

I've not ever been taught it in class

but it's literally just

substitution

so it's fine

easy stuff

I got the second part done easy cuz it's just plug and chug of part one

[marsh mountain]

yep

[oblique estuary]

I have 3 questions left but I'm gonna go to my prof's office hours

since I have a bunch of other questions relating to other stuff

she wants us using this AI thingy that I opted out of because I didn't want to be a part of a research study

it's not supposed to affect our grade though opting out of the study

so I'm gonna ask her

[marsh mountain]

good luck!

[oblique estuary]

thanks!

have a good day!

.close

[final saddle]

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[marsh mountain]

you too!

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[obsidian knot]

.close

!close

[small grail]

it is closed

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[cinder galleon]

Hello, I need help with trying to prove an equation from James Stewart's Review of Algebra

[cinder galleon]

First, I'm gonna show the documentation because it's available for free on the james stewart website

https://stewartcalculus.com/_update/media/common/reviewofalgebra.pdf

So, in this file there's a thing called the Distribution Law

Actually it's Distributive Law, my bad

The file says that if we use the distributive law three times, we get this result:

The thing is, how I would use the Distributive Law three times by myself? I'm really confused on this part. If I just keep multiplying the letter A with (b + c) three times, it doesn't get the result from the text book. I got a^4b + a^4c, which makes no sense.

[onyx peak]

a, b, and c here are any terms, they dont need to be the variables a, b, c themselves

for example that law can be used to show that
4(3 + 5) = 4*3 + 4*5

or that (a+b)(c+d) = (a+b)c + (a+b)d

in the second case, the "a" in the distributive law is (a+b), the "b" is c and the "c" is d

look at the color coding here

red(blue + green) = red*blue + red*green

its the same law in both cases, although it uses slightly different letters / terms

[cinder galleon]

if i wanted to use the distributive law two times instead of three, how it would look like? using the letter terms

[onyx peak]

this is what we got after one application

now the 2 other applications are used to expand (a+b)c and (a+b)d

you technically need commutativity as well, to convert it to
c(a+b) and d(a+b)

but now, you can expand it using distributivity

note that the terms circled in purple are the same

i can expand c(a+b) usind distributive law to ca + cb

then i can similarly expand d(a+b) to da + db

that's 2 other applications of the distributive law

so (a+b)(c+d) = (a+b)c + (a+b)d (Applying distributivity here)
= c(a+b) + d(a+b) (Commutativity)
= ac + bc + d(a+b) (Applying distributive law on c(a+b))
= ac + bc + da + db (Applying distributive law on d(a+b))

so total of 3 applications of the law get us to the result

• 2 minor applications of commutativity

[hot hornet]

yeah commutativity is key here

[onyx peak]

the most important thing to realize when reading laws like this one is that a, b and c dont have to be just the letters a, b, c

if any expression is of the pattern
something1(something2 + something3)
you can make it
something1 * something2 + something1 * something3

it doesnt matter what something1, something2 and something3 are

the only thing that matters is the pattern
# (# + #)

[hot hornet]

yes, that's why flowchart diagrams can be helpful for this

[final saddle]

@cinder galleon Has your question been resolved?

[cinder galleon]

i'm trying to understand 😭

idk what to choose

i can do this with numbers but it gets abstract with letters

[onyx peak]

# (# + #)
you need to spot a pattern like this basically

or (# + #) #

it'll be clear once you do enough exercises

https://www.khanacademy.org/math/algebra-basics/alg-basics-algebraic-expressions/alg-basics-distributive-property/v/distributive-property-with-variables-exercise

[cinder galleon]

I actually did the review exercises, but i'm trying to prove it on the letters to see if this makes me better at math

I think it's called the proving method I think

Basically you prove the equations on the book to make sure you know it?

[onyx peak]

yeah, ig thats a good way to learn it, but maybe it would be worth it to try it with slightly simpler combinations of letters first

and not with that double parenthesis (a+b)(c+d) thing

[cinder galleon]

when i see something like x (y + z) i multiply the X with Y and Z, so it would result in xy + xz

[onyx peak]

yep, thats how it works

and when you see
(a+b)(y+z), you can just interpret (a+b) to be ur x and do the same

(a+b)y + (a+b)z

[final saddle]

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[slate anchor]

We consider positive integers of the form
$$
aabb = 1000a + 100a + 10b + b,
$$
where $$a \ne 0$$ and $$a, b$$ are digits.

(a) Prove that every integer that can be written in the above form
is a multiple of $$11$$.

(b) Determine all integers of the above form that are multiples of
$$4$$ and $$9$$, and write each of them as a product of prime factors.

[soft zealot]

ponkur

[trail mango]

which part are you working on

well actually i will disappear for a little bit. if the first part, just try to write aabb as 11k for some integer k. if the second part, maybe look at multiples of 4 and 9 and figure out what they “look like”

[final saddle]

@slate anchor Has your question been resolved?

[slate anchor]

the b

ik how to get nummber weith can multiplye with 4,9

just the whole b

i need some help i got the a pretty easily

[final saddle]

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[toxic tree]

Before we even focus on the asymptotes im trying to get the domain

I set the denom != 0 and tried to solve but got stuck

sqrt(x^2 + 3x) != -x
x^2 + 3x != x^2

Then one of the terms dissapears

3x != x^2 - x^2
3x != 0
So we just have x != 0 which cant be right

We need 2 values for x

So what did I do wrong?

[steep nest]

well lets just say, when do we have 1/0?

[toxic tree]

When the denom = 0

[toxic tree]

Which is why I tried this

But ya idk what I did wrong

[modest badge]

there is another condition for the domain

the inside of the square root cannot be negative

[toxic tree]

I dont think it was at any point in my work right?

Unless you mean squaring the -x to turn it into x^2

[modest badge]

setting the denominator to 0 helps to find vertical asymptotes

[winter lava]

there's nothing wrong with your work

you're right that x=0 is not in the domain

[toxic tree]

But there should be another right?

[modest badge]

but there can be other reasons why a region of the x axis is excluded from the domain, one of which is square roots of negatives

[toxic tree]

Cause the highest value of the denom is x^2

[winter lava]

yeah it's like olivas said

[modest badge]

that x^2 is inside a square root, so the highest degree is not that

[winter lava]

check x^2+3x >= 0

[toxic tree]

Ohhhhhhhhhhhhh

I see what you guys are saying

So x =! 0 and x >= -3

[winter lava]

or

no

x^2 + 3x >= 0 means
x(x+3) >= 0, so there's cases

[modest badge]

close, but you're mixing the sign

[toxic tree]

Wait how sry

x^2 + 3x >= 0

x + 3 >= 0

[winter lava]

you can't divide by x

[toxic tree]

x >= -3

But we already know x != 0

So I dont need to worry about that factor right

[winter lava]

use the fact that ab is positive when a,b are both positive, or both negative

[modest badge]

it should be x =< -3

if x is between -3 and 0, x is negative, but (x+3) is positive, so x(x+3) is negative, and we can't have that

for both x and (x+3) to be negative, x=< -3

[toxic tree]

Wait first before any of that why cant I divide by x sorry

[winter lava]

because dividing by a negative flips the direction of the inequality

but dividing by a positive doesn't

you don't know if x is positive or negative

[toxic tree]

Ahhhhh

Okokok

[toxic tree]

How do we know (x+3) is positive sry

[modest badge]

because -2 + 3 > 0, for example

in the case x is between -3 and 0

[toxic tree]

Ahhhhhhhhhhhhhhhhhhh

Okokokokokokok

[toxic tree]

Alright I understand your point now

Ok I get it

Thank you so much guys!!!

❤️

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[lapis wedge]

is the reason we need to have A intersect U not be empty as they said at the bottom is because it would make the above definition vacuously true if it were empty?

[granite thorn]

U cap A \ x_0 cannot be empty since x_0 is defined as a limit point

[lapis wedge]

yeah

[granite thorn]

Then what is it ?

[lapis wedge]

i’m saying do we define x0 to be a limit point to prevent the definition at the top from being vacuously true

[granite thorn]

Yes

[lapis wedge]

ok thx

[granite thorn]

One of the reason the convention is this way i think is because you want to use limits to carracterise topological properties sometimes

But ye just suppressing the useless case is also good in general i guess

[lapis wedge]

hm I see

thanks

.solved

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[halcyon knot]

hello

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@halcyon knot Has your question been resolved?

[final saddle]

Channel closed

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[latent dragon]

<@&268886789983436800> troll

[faint edge]

Don't troll here

[opaque ember]

@pallid sentinel pls do not troll

[steep hatch]

mod party

[opaque ember]

.close

[final saddle]

Channel closed

Closed by @opaque ember

Use .reopen if this was a mistake.

[latent dragon]

free robux scam in bio too.

[steep hatch]

good catch thanks

[latent dragon]

no problemo.

[final saddle]

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.
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[boreal smelt]

how to solve this?

[woven ledge]

check the minimum and maximum values for the lhs and rhs

[boreal smelt]

why?

[woven ledge]

obviously you cant solve for x and y normally

so its gonna be some trick like that

but check the minimum and maximum values and see what you get

[boreal smelt]

alrightt

the minimum values of both the functions are same

i.e. 13

[woven ledge]

not exactly

[boreal smelt]

sorry maximum

[woven ledge]

the maximum of the lhs is the minimum of the rhs

yes

so both are equal to 13

[boreal smelt]

yea

[woven ledge]

can you solve for x and y now

[boreal smelt]

yea

how do I find out x?

help

😭

@woven ledge

[woven ledge]

do you know the formula for acos x + b sin x

[boreal smelt]

of range?

[woven ledge]

no just like a kind of simplification

[boreal smelt]

no

[woven ledge]

ok so let $a = r cos y$ and $b = r sin y$

[soft zealot]

CherryMan

[woven ledge]

then $a \cos x + b \sin x = r \cos y \cos x + r \sin y \sin x$

[soft zealot]

CherryMan

[woven ledge]

$= r cos(x-y)$

[soft zealot]

CherryMan

[woven ledge]

if we solve for r and y we get $r = \sqrt(a^2+b^2)$ and $y = \arctan (\frac{b}{a})$

[soft zealot]

CherryMan

[woven ledge]

so in the end $a \cos x + b \sin x = \sqrt{a^2+b^2}\cos(x - \arctan(\frac{b}{a}))$

[soft zealot]

CherryMan

[woven ledge]

did you understand the process?

[boreal smelt]

yea

[woven ledge]

so if we want to solve for this we need $\cos(x-\arctan(\frac{12}{5})) = 1$

[soft zealot]

CherryMan

[boreal smelt]

yea

[woven ledge]

that means the argument is (2n+1)pi

but if we just take one solution where the argument is 0

[boreal smelt]

yea

[woven ledge]

then we get $x=\arctan(\frac{12}{5})$

[soft zealot]

CherryMan

[woven ledge]

and im sure you got that y=2

put it into 12cot(xy/2) and see what you get

[boreal smelt]

thankssss

[woven ledge]

welcome

[boreal smelt]

.close

[final saddle]

Channel closed

Closed by @boreal smelt

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[weary edge]

Can someone help here

I saw someone said that angle EAF is 15

So angle ECF is 30

But im just confused how they knew it was 30?

[small grail]

,rccw