help-26

[thorny flame]

Veni, vidi, perii

[prisma mesa]

Isn't this the definition?

[ivory sorrel]

yeah, but I want to prove that |1/x| = 1/|x|

[prisma mesa]

How does it even follow from this?

[ivory sorrel]

oops

I think It might be better to prove |a||b|=|ab| first

[neon iron]

If you have the lemma |ab|=|a|.|b| then put a = 1/x and b=x

[ivory sorrel]

I mean that was the previous question

[prisma mesa]

cool, then apply it

[ivory sorrel]

okay, so |1 \cdot 1/x| = |1| |1/x|

[pearl fog]

technichally, what is this study? its kinda intresting on how you have a massive open field but here you decided to go to the forest, shoot yourself in the leg, blind yourself and see how it goes

[prisma mesa]

yeah, which is just saying |1/x| = |1/x|

[prisma mesa]

this sub will be better

[ivory sorrel]

LOL

Let me ponder over this a bit more

hmm

[prisma mesa]

I wonder what would you call using only FOL axioms and ZFC then

[pearl fog]

chopping off your limbs.

(mods dont ban me 🙏)

[ivory sorrel]

ok, yeah. makes sense

a=|1|

b=1/x

[prisma mesa]

this one works too

wait it doesnt

hmm

[prisma mesa]

this would give |1/|x|| = |1/|x||

[prisma mesa]

this sub is the best you can do probably

[soft oyster]

it should follow from this

$$ |x| =
\begin{cases}
x & \text{if } x \geq 0,
-x & \text{if } x < 0.
\end{cases} $$

[ivory sorrel]

b=1

right

[thorny flame]

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

[soft oyster]

woops

[prisma mesa]

or you could go completely without that lemma, and do it by cases

[prisma mesa]

if b = 1, then it will always just give |a| = |a|

b cant ever be 1

[soft oyster]

but yeah it should follow from the piecewise definetion of absolute value

[prisma mesa]

unless you want a completley useless result

[ivory sorrel]

why

[soft oyster]

subsitute 1/x for x

[prisma mesa]

|ab| = |a| * |b|
|1 * a| = |a| * 1
|a| = |a|

this is what happens when you plug in b = 1

try a = x and b = 1/x

or just do it without this lemma and prove it for x > 0 and x < 0 seperately

[topaz sinew]

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[spiral yoke]

pro

[spiral yoke]

help me 😢😢 idk where to start

[sour bloom]

are there any angles which are the same?

[spiral yoke]

i dont think so

[sour bloom]

check every angle in the diagram

two important angles are the same

[spiral yoke]

why do u think so

32 is the same as angle A or what

;-;

[mint hare]

you can use that AB and DC are parallel

[spiral yoke]

o

so is angle dba = angle bdc?

@mint hare

[mint hare]

You mean at D and at A?

oh whoops you mean at D and at B

[spiral yoke]

what

yes

am i wrong

[mint hare]

You're right

[spiral yoke]

since its alternate

[mint hare]

ABD = BDC

yes

[spiral yoke]

omg im smart

tys,

tysm

so for that i can use sine rule to get th value of angle 0?

[mint hare]

yeah

[topaz sinew]

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[elfin niche]

So I got 3 points

I'm assuming that vertex is (0,0)

(-200, 100) left tower
(200, 100) right tower

[inner wren]

where is the road?

oh the surface of the bridge

nvm

Can you clarify what is meant by the 'road'?

[topaz sinew]

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[ripe maple]

Calculating my reactionspeed and am not sure if this is correct, i used desmos to calculate but the reactionspeed of 0.019 seconds doesnt seem reasonable: t= speed in seconds) t = √2 × 0,136/9,82 ≈ 0.019 is this correct or did i mess up? any help very appreciated since i need this for a paper im writing for school, ty in advance 😄

[restive inlet]

where are those values coming from?

[inner wren]

!original

[topaz sinew]

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.

[ripe maple]

gimme 1 sec js gna type it all out

[inner wren]

ok

[ripe maple]

started with s=½gt² (weightaccelleration equation) s is time, g is weightacceleration on the surface of the earth and t is time

then solve for time

t= √2s/g

then i inserted the object falling length 13.6 cm = 0.136m

[restive inlet]

you neglected order of operations when entering the value into the system

to be clear, you should have
$$t = \sqrt{\frac{2s}{g}}$$
$$t = \sqrt{\frac{2\cdot 0.136}{9.82}}$$

[thorny flame]

ℝαμOmeganato5

[restive inlet]

also why 9.82 instead of 9.81 for gravity

[ripe maple]

noway i got gravity wrong xd

it was in my textbook i think

okay i got it to 0.166 seconds that makes more sense ty omega ❤️

$$t = \sqrt{\frac{2\cdot 0.136}{9.81}}$$

[thorny flame]

𝖘𝖆𝖓𝖉𝖈4𝖘𝖙𝖑𝖊

[ripe maple]

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[final fossil]

how can i isolate lambda here

[neon iron]

in terms of b>

?*

[final fossil]

yes

can i log linearise?

[neon iron]

nope

log(a+b) doesnt equal loga+logb

wait

nah u cant isolate here

[final fossil]

😦

[neon iron]

you can isolate but then youll get it in terms of dy/dlambda

again y is a new variable so nope

[final fossil]

k ty then

[inner wren]

just find the value of lamba?

wait

nvm

those both expressions are the same

[final fossil]

yea

[inner wren]

oh that was your process

[final fossil]

its part of a lagrangian function

im trying to solve lambda in terms of b

then find optima

[neon iron]

ive never heard either of those terms before what 😭

lagrangian function??

[neon iron]

anyways ty guys for you work

.solved

[topaz sinew]

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[short trail]

I need help with something sadly

[short trail]

Is there a way to write this easier?

I've tried to cancel the cos(30) and sin(45) out, but it appears I was wrong

Ohhh wait I forgot sin(45) also multiplies with sin(30)

[odd flower]

ngl

its easy

just brute force it

dont waste ur time thinking for shortcuts

[short trail]

Yeah, but writing that in a calculator during an exam isn't so good

[odd flower]

tfu need a calculator

[deep crow]

Sin(45) * 1/sin(45) cancels

Simplify that and then use unit circle

[odd flower]

is a + sign

[icy sky]

alternately, divide by sin(30) in the numerator and denominator and move the cos over to get cot double angle iirc

thats unspottable

cot angle sum*

[short trail]

I can't type cot in calculator

[icy sky]

oh it doesnt even matter because thats cot75

id jst brute force it, those are nice enough values

[short trail]

Well rip my time in exam, ty then

[icy sky]

gotta remember your first quadrant unit circle values

[odd flower]

or

u could just combine the bottom fraction

and watch everything cancle

[short trail]

I have tried to cancel things out, but I get different results

Actually I might have been just dumb

Oh yeah I was

cos(30)/3+cos(30) is not 1/3+1

It's fine then thank you I just made a no brainer

.close

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[wintry urchin]

Is this a valid proof?

[wintry urchin]

I meant "linearly independent" on the third last line

[neon iron]

I don't know what plus/minus theorem is, but I guess it is just computation. Yes, this argument will work, and you can also argue that this algorithm will eventually halt as the vector space is finite dimensional.

[wintry urchin]

👍 Thank you

.close

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[mortal marten]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mortal marten]

1

[topaz sinew]

@mortal marten Has your question been resolved?

[mortal marten]

Thanks for repeating the question idk how to read

Are you gonna help or what

Exactly

hows BC = BE + EC

Oh

Yeah i see

Hold on let me write allat

Ok ok

yeah

Ok ok im writing hold on a sec

could you explain this

Afe isnt a triangle tho?

AFE isnt a triangle is it?

Did you make it a triangle

ohhhh

Yeah

so all ratios are equal?

alright i will try to solve now but i wont close channel incase i dont figure it out

The angle-angle of AFE and ADC are <EAF=<CAD and what else?

The angles F and D? how are they equal

Ah ok

Hey i still cant get to the solution

wait can you explain this too?

[topaz sinew]

@mortal marten Has your question been resolved?

[mortal marten]

<@&286206848099549185> can yall help? read the previous messages

[mortal marten]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mortal marten]

2

[topaz sinew]

@mortal marten Has your question been resolved?

[topaz sinew]

@mortal marten Has your question been resolved?

[mortal marten]

.close

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[north lion]

am i doing anything wrong? it’s giving different answers than expected but i can’t find the mistake

[scenic flare]

what's the expected answer

I don't see an error with the diff so far

[north lion]

i don’t know, i just plotted the function in desmos and when i graphed “f’(x)” the function looked different than when i plotted the derivative i computed

not that i took a low upper limit for the sum, i tried n=500

[sweet shard]

screenshot your graphs

[scenic flare]

hm check the symbols and send the desmos plot

[north lion]

gonna take a bit bc i closed the app

[scenic flare]

could be that desmos doesn't handle gamma,digamma,inf sums that well, I haven't used it for that, but maybe it's reliable enough

[north lion]

i can’t find how to share desmos graphs on mobile

[whole geode]

That looks the same to me

The difference between the two is likely a difference in the convergence speed of the summations

[north lion]

look around x=0

[north lion]

okay

but at a limit that high it shouldnt be almost a quarter unit right?

[whole geode]

It could very well be the case, and I would assume it is unless you can put rigorous bounds on the error

[north lion]

also, this sum came up when calculating $\int_{0}^{\infty}\tfrac{\ln(x)}{\cosh(x)}dx$, but when i calculated f’(0) (the value of the integral) it was almost the same (like 5 decimals or smth)

[thorny flame]

pixel

[north lion]

i’ll close this channel now thanks for the help guys

.close

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[tender abyss]

no idea where to start from please help

[topaz sinew]

@tender abyss Has your question been resolved?

[tender abyss]

<@&286206848099549185>

<@&286206848099549185>

[twin heath]

ew

[twin heath]

okay so first of all, what relation do f(x) and g(x) roughly have

[tender abyss]

g(x) is a compression of f(x)?

and a reflection

@twin heath

[twin heath]

sure

let's try to make those a little more precise

first of all

pick any point of f(x) to try and "match up" with the bottom graph

say..

[tender abyss]

-8,9 becoemws 7.4

7,4

[twin heath]

it's -7, 9 but

close enough

[tender abyss]

misaw mb

[twin heath]

let's try this part since it looks a little easier to deal with

first of all

it's very clearly the same part as g(x)'s leftmost

segment

[tender abyss]

mhm

whats after

[twin heath]

next, we can also very clearly see that it goes only (about) half as high as f(x)

[tender abyss]

so the comporesions is 1/2

[twin heath]

i think it's closer to 4/7 but im going to want like

a little less than a lowball

wait

i miscounted

it's more like.. 2/3rds

actually wtf am i cooking

💀

let me count one more time lmfao

yeah it's 1/2 okay

ignore that

i can count

trust

so right now we have 1/2f(-x)

[tender abyss]

mhm

[twin heath]

now we can finally

work on the vertical shift

[tender abyss]

-2?

[twin heath]

lot more than -2 it looks like

lets pick f(2) to test this on

aside from the vertical shift

we've got

[tender abyss]

if tis 4 it reflects to -4 and -2 is -6

[twin heath]

he might be cooking

[tender abyss]

but on the key it syas -17/2

[twin heath]

oh yeah so

lets use the horizontal line instead

[tender abyss]

mk

[twin heath]

so

we've got the red line on the left as

1/2f(-2)

which is pretty close to g(x), but we just need to shift it

its y value is 2.5, g(x) is -6

so we get -8.5, or (guess what)

-17/2

!!!

[tender abyss]

wdym y value?

[twin heath]

okay so

you take the y value of f(x) at.. sayyy

2

which is 5

right

[tender abyss]

mhm

[twin heath]

that's the y value for f(-2)

now we just divide that by two for 1/2f(-2), which was the equation we came up with

[tender abyss]

mk

[twin heath]

the y coordinate of that point is -2.5

[tender abyss]

wait

-2/ 1/2 is -4 -2 is -6

[twin heath]

whar

[tender abyss]

how 2.5

2/2 is 1

[twin heath]

f(2) is 5

as in

at x=2, f(x) is 5

[tender abyss]

ah i misread som

mk

[twin heath]

no it's okay

[tender abyss]

2.5

[twin heath]

g(x) is -6 at x=-2

so

we just need to move it down

2.5 - what = -6

[tender abyss]

3.5

wait

[twin heath]

waiting

[tender abyss]

7.5

[twin heath]

still nope

[twin heath]

that's 1.5

[tender abyss]

8.5

[twin heath]

there we go

or

if you make it a fraction

17/2

[tender abyss]

ahhh

mk

[twin heath]

so your final answer is

$\frac{f(-x)-17}{2}$

[thorny flame]

Serphic

[tender abyss]

mk

tysm

[twin heath]

np!

[tender abyss]

.close

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[topaz robin]

hi i have no clue where to start with this questions

[thorny flame]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 6)

[topaz robin]

what does that mean ^ 🧍‍♀️

[flint rock]

Can Antoine help me

Anyone ^

With genotry

[wary tulip]

just think about playing one round first (so each person rolls once)

what's the probability A gets eliminated? what about A doesn't get eliminated but B does?

[topaz robin]

so like Pr(A is elim)*Pr(b and c don't get elim) ???

[wary tulip]

hm just pretend the "round" ends if one gets eliminated

[topaz robin]

sorry huh 🧍‍♀️

[wary tulip]

🧍🏻‍♀️

ok like

there are 3 different ways we can go down to 2 players

A,B left, A,C left, or B,C left

well nvm that

we can just think of it like

we play a round where every player rolls once and the game ends immediately if one player gets a 6 during the round

but if no one rolls a 6 we're just gonna do the same thing again

so we can just think about one round first

[topaz robin]

there is this equation that they gave us but i have never seen this in my life HELPAHSD

wait lemme show

[wary tulip]

[topaz robin]

in round 1 would the probaility of each one of them getting elim be just 1/6 or have i got this completely wrong

[wary tulip]

not quite

for A it is 1/6

for B to be eliminated, we'd need A to not roll a 6, then B to roll a 6

[topaz robin]

so 5/6 * 1/6?

[wary tulip]

yea

[topaz robin]

and then c 5/6 * 5/6 * 1/6 ?

[wary tulip]

yea

[topaz robin]

and now i have no clue where to go from here man :D

[wary tulip]

soooo

these are the probabilities they get eliminated
A 36/206
B 30/206
C 25/206

[topaz robin]

how tf

[wary tulip]

in one round

[topaz robin]

how did u get those 😭

[wary tulip]

5/6 * 5/6 * 1/6 = 25/206

1/6 = 6^2/6^3 = 36/206

i just gave them all the same denominator

[topaz robin]

OH

OKAY

now what? HAHAJKSDHKJ

[wary tulip]

hmm how to explain this

are you familiar with conditional probability

[topaz robin]

yeah i would say i'm alright with it

[wary tulip]

ok

we can just assume we're on the round someone gets eliminated

the ones before it (if any) where nobody gets eliminated don't matter

so then you can compute the probability A gets eliminated given that someone is eliminated (i.e., at least one person rolls a 6)

and same for B, and C

[topaz robin]

wait sorry let me try take in everything u said and try the question again 😭 thank u !!

.close

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[languid depot]

Help

[topaz sinew]

Please don't occupy multiple help channels.

[viral fulcrum]

just send the question

and use only 1 channel

[sharp dew]

Only you can help you

Do you wish us to check the answer

[languid depot]

Yes

Bc=cd is what i picked but idk know if theres another answer

[sharp dew]

Does the answer matter? Like if you handed it wrong, what’s the consequence

[languid depot]

Bad grade..

[obtuse parrot]

hiiiiiiii

[sharp dew]

You should check it yourself, grade doesn’t really matter tho. The grade will come after should you put the efforts in

[craggy haven]

this is supremely unhelpful. if you don't have anything constructive to say it would be better to just not say anything.

[sharp dew]

I apologize

[neon iron]

thank you for reading my modmail

sorry if my ton of messages (separated) were distracting

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@languid depot Has your question been resolved?

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[gilded trench]

How do I do this. Just number five

[astral blaze]

polynomial with degree n should have n not necessary distinct solutions

but complex solution is special because if one is the solution, its conjugate is also the solution

Because you can always create a polynomial with degree n using multiplication
$P(x) = a(x-r_1)(x-r_2)\dots (x-r_n)$
for constant $a$ and solutions $r_1,\dots r_n$. Notice that this multiplication results in $ax^n + \dots$ polynomial

[thorny flame]

annoying

[gilded trench]

Yes

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[stark mica]

How to do 9

[neon iron]

,rotate

[neon iron]

which question

[vocal escarp]

Matthew my boy

9th question

[sweet shard]

Draw a triangle with hypotenuse 7 and one side 3. Find the angles corresponding do your inverse sine/cosine values

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[eternal nymph]

i am confused as to how this is wrong. this is my calc 3 class and i really dont understand

[gleaming reef]

show us ur work

[alpine mist]

,rccw

[alpine mist]

shouldn't it be -55?

[vernal matrix]

(while they accepted it, probably better to use the (1, -8) one tbf)

[eternal nymph]

oh my gosh i was losing my mind and it was a negative sign thank you lol

[vernal matrix]

Seeing in the first part you find the slopes, call them m, from there you can write the parallel vectors as (1, m)

[eternal nymph]

okk

thank u guys!

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[crisp wren]

How do you find angle Theta? I don't understand how to even after watching multiple videos. I feel like I'm missing something.

[half gazelle]

First of all, you have cosine squared, to find angle you have to solve this equation for cos first

Any ideas how to do it?

[crisp wren]

Not really

[half gazelle]

Hmm

What if we use substitution?

Like

[crisp wren]

I have the answer, just need to know how to get it.

[icy sky]

i would start by square rooting both sides

[half gazelle]

cos(theta) = t

[half gazelle]

Yeah I was trying to say this but without saying this

🥸

[icy sky]

why not just say it

[crisp wren]

Here is the answer I'm trying to understand how to get.

[icy sky]

did you square root both sides dottoe?

[half gazelle]

Bc there is no point on giving him solution if he will not understand, I wanted him to solve this, not me

Idk

Maybe I'm wrong

[crisp wren]

I watch videos about it, but it keeps talking about other equations I don't know about.

[icy sky]

how about you try square rooting both sides

[crisp wren]

I'm trying to.

I was mainly just asking if there was a equation wheel that I needed to know about.

[onyx summit]

Probably about arccos

You need it to solve this

[crisp wren]

ok, ty.

[onyx summit]

Its the inverse function of cosine

[crisp wren]

Ah, i see, now I understand.

That answered my question, thank you.

[onyx summit]

Also make sure you can get all 4 answers

[crisp wren]

I was just having trouble with figuring out how to solve for 0 in cos(0).

[onyx summit]

If you need a hint for why there are 4 answers you can ask, otherwise I think you can figure it out on your own

[crisp wren]

I may need a hint, that is the main part that confused me. I know it has something to do with quadrants.

[icy sky]

what happens when you square root both sides

[crisp wren]

1/4 becomes 1/2, having only Cos(0) = 1/2

[icy sky]

are you sure thats the only possible square root?

[crisp wren]

also -1/2

[icy sky]

how many solutions does cos(theta)=1/2 have? (in the interval [0,2pi))

[crisp wren]

2

[icy sky]

what about cos(theta)=-1/2?

[crisp wren]

2

[icy sky]

how many solutions in total?

[crisp wren]

4

I was just having trouble with how it went from 1/2 to 2pi/3

[onyx summit]

two more because cos(2pi-x)=cos(x)

[crisp wren]

Ah, it was radians/accros that I was having trouble with.

[onyx summit]

From 1/2 to pi/3 is because of arccos, and then the other angle that works for that is 2pi-pi/3, which is 5pi/3, due to cos(2pi-x)=cos(x)

You can do the same process for -1/2

[crisp wren]

Oh yeah, I remember now, the unit circle.

Ok, I remember now, thank you very much.

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[ashen frigate]

Bark vark

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[cerulean star]

Incoming question re: Peano addition. I'm just now learning it, and am getting hung up on one of the steps in a simple introductory problem.

[cerulean star]

Here is the original problem and the provided workthrough and solution:

[prisma mesa]

S(S(S(S(S(0)) + 0)))

So you dont know how to proceed from here?

S(S(S(S(S(0)) + 0)))
S(S(0)) + 0 = S(S(0))
S(S(S(S(S(0)) + 0))) = S(S(S(S(S(0)))))

[cerulean star]

No, the step prior to that. I don't understand how you get from S(S(S(S(0))+S(0))) to S(S(S(S(S(0)+0)))). I keep ending up with something different.

[prisma mesa]

S(S(S(S(0))+S(0))) = S(S(S(S(0) + S(0)))) = S(S(S(S(0 + S(0))))) = S(S(S(S(S(0)))))

@cerulean star this should work I think

it's just moving the S(0) inside

[cerulean star]

Okay, I'll check it out. Lecture starting here in a minute, so I'll close the channel and come back to the server after I've tried the problem again. Thanks!

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[uneven glen]

can tan^2(theta) = sin^2(theta)/cos^2(theta)?

[topaz sinew]

Please don't occupy multiple help channels.

[uneven glen]

ok

[shut night]

yes

[sonic escarp]

type .close

[cunning snow]

can? It is equal to that

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[sand stream]

How would i differentiate this with respect to x without a calculator on tech free

[sand stream]

The textbook got

But this didn't show me any working out so i was unsure how it did that like how is it 24(3x+4)^3 should it not be 8(3x+4)^3

[lusty niche]

They used chainrule

U do normal power rule

But multiply buy the derivative of the inside

The derivative of 3x+4 is just 3

So multiply by 3

You would use chain rule when you have a function within a function

[sand stream]

Oh ye i remember now

Yup thank you soo much

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[outer mulch]

Im doing number 6

Could anyone check that I'm correct

Im not confident in my answer

<@&286206848099549185>

[lusty niche]

Replace the x with 4x+9

And then jst expand

[outer mulch]

So would what I have be correct

Because i did the steps

[lusty niche]

f(4x+9)= -(4x+9)/2 - 3

Expand from there

[torpid sparrow]

I don’t think that is what is being asked

[lusty niche]

Oh wait whats the q asking for then

[outer mulch]

I got -2x-12

And then add 9

So -2x-3

But I don't know how to write it

[torpid sparrow]

What do you mean how to write it?

[outer mulch]

Like in the answer

Would I write it as f(-2x-3)

Or some other way

[torpid sparrow]

No

[outer mulch]

So I have it right?

[torpid sparrow]

No

[outer mulch]

What am I doing wrong?

[lusty niche]

No, f(x) is jst the notation for a function meaning that whatever u input into x, it will give u an output so e.g f(x)=x+1 thats simply saying whatever is input into x, add one and thats the output, think of functions as a machine that does a certain process to whatever value you input into it, to give you an output

[outer mulch]

Oh ok

So I don't include the f

I just have -2x-3

Because that is what is outputted

[lusty niche]

Wait

[lusty niche]

Yh

[torpid sparrow]

Recheck your numbers

[outer mulch]

For who?

How is it this?

Ok thank you

I have to go know

Now

Bye

[lusty niche]

Answer is (-4x-15)/2

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[rocky jewel]

As for the values that this function can take

is geogebra messing with me?

Is it not ]0, 1]??

Or does it not care about the fact that we have a negative square root since arctan squares it? ._.

Yeah No seems lite it’s the opposite of What i thought lmao

[agile harness]

x≠0

[rocky jewel]

Yeah hence the ]0

[agile harness]

use (

to indicate an open bracket

[rocky jewel]

But isn’t the restriction on the f(x) based on the g(x)?😭

interesting, we use ]

[0 if included ]0 if not

[agile harness]

well arctan is defined for all x

so it should only be the sqrt part

[rocky jewel]

Yes but square root only for positive?

[agile harness]

ahh true

[rocky jewel]

And unless i’m too dumb to me it seems like it’s when 0<x<=1

[agile harness]

because it is

[rocky jewel]

so What should the domain for the entire function be then?

[agile harness]

,w graph arctan((1-x)/x)

[thorny flame]

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[agile harness]

oops forgot the sqrt

,w graph arctan(sqrt((1-x)/x))

[thorny flame]

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[rocky jewel]

oh so then i was right?

[agile harness]

yes

[rocky jewel]

Oh mah gawd

I just might pass my exam

[agile harness]

you probably typed it in wrong

[rocky jewel]

Ok ok Wait since you know How that stuff work

What’s the derivative

-1/(x*2sqrt((1-x)/x)))?

pleaselordletthisbethecase

[agile harness]

derivative of arctan(u) is $\frac{du}{1+u^2}$

[thorny flame]

knief

[agile harness]

differentiating the sqrt function to find du

gives

well first i’ll rewrite the inside of it because i don’t feel like doing quotient rule

you’ll get (1/x)-1

hmm

[thorny flame]

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

[agile harness]

hmm

i’m new to latex i’ll just type it

you know how to differentiate though right

[rocky jewel]

Well i hope so

But Is it What i wrote?🤯

[agile harness]

,w differentiate (sqrt((1-x)/x))

[rocky jewel]

oh no

[thorny flame]

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[agile harness]

then that’s your du

,w differentiate arctan(sqrt((1-x)/x))

[thorny flame]

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[agile harness]

this is your final answer

[rocky jewel]

-1/(x*2sqrt((1-x)/x)))?

Does look so

fuck ye

Okok and then

Ln(tan^2(x)

is that 2/sinxcosx?

[agile harness]

cot^2(x)(2tan(x))(sec^2(x))= 2tan(x)csc^2(x)

[rocky jewel]

ruhroh

[agile harness]

which is 2/sinxcosx

[rocky jewel]

RUH WOOOH

okok and um

[agile harness]

i have to go

[rocky jewel]

Haha Aight

[agile harness]

i have class

[rocky jewel]

Thanks mage

[agile harness]

you’re welcome

[rocky jewel]

No problem, glgl

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[strong sable]

Find the area of the shape bounded by the curve ( y = e^{-x} \sin x ) (for ( x \geq 0 ) ) and the x-axis.

[thorny flame]

miyo

[strong sable]

! occupied

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[sand oasis]

Oops

Sorry

[strong sable]

haha it's ok

[sand oasis]

Ok bye lol

Ignor eit

[strong sable]

you can just delete it

[topaz sinew]

@strong sable Has your question been resolved?

[north lion]

you have learned integrals already right?

if it's asking for the area bounded by a curve and the x-axis, it's just the integral of that curve

since it's asking x>=0, we need to have limits from x=0 to infinity

[strong sable]

yes This is what i've done: ∫ e^(-x) sin(x) dx = -e^(-x) (sin(x) + cos(x))

[north lion]

could you explain what you've done here?

[strong sable]

integration by parts

[north lion]

yes, that's the right move, but why do you have another integral on the right side?

after integrating by parts you should have: something -integral(something)

[north lion]

yes

that's correct

[wet harbor]

how is last step correct

[north lion]

oh yeah i overlooked that

[wet harbor]

xD

it's very close though

[north lion]

you forgot the integral part of second time integrating by parts

[strong sable]

$\int e^{-x} \cos x , dx = -e^{-x} \cos x - \int (-e^{-x})(-\sin x) , dx = -e^{-x} \cos x - \int e^{-x} \sin x , dx.$

[thorny flame]

miyo

[north lion]

yes, very good, now sub that in here

[wet harbor]

i don't think this is right

he went backwards

first he integrated cos and differentiated the exp

now he differentiated the cos and integrated the exp

[north lion]

oh yeah because then we just get zero

[wet harbor]

after substitution he will get back $\int e^{-x}\sin(x)dx = \int e^{-x}\sin(x)dx$

[north lion]

yes exactly

[thorny flame]

Mohamed Mohsen

[north lion]

so we need to take u=e^-x and dv=cosx the second time integrating by parts

[strong sable]

My no brain moment lemme re calculate

[strong sable]

[ \int e^{-x} \sin x , dx = -e^{-x} \sin x - \int -e^{-x} \cos x , dx ]
[ \int e^{-x} \sin x , dx = -e^{-x} \sin x + \int e^{-x} \cos x , dx ]
Where [\int e^{-x} \cos x , dx = -e^{-x} \cos x - \int e^{-x} \sin x , dx ]
[ \int e^{-x} \sin x , dx = -\frac{e^{-x} (\sin x + \cos x)}{2} + C ]
[A = \int_{0}^{\infty} e^{-x} \sin x , dx ]

[north lion]

yes, thats correct

fr this time i know earlier i said it was correct but wasn't

so now just plug in the limits of integration

[thorny flame]

miyo

[strong sable]

[\lim_{x \to \infty} -\frac{e^{-x} (\sin x + \cos x)}{2} = 0 ]
When x approaches 0 [-\frac{e^{0} (\sin 0 + \cos 0)}{2} = -\frac{1 \cdot 1}{2} = -\frac{1}{2} ]

[north lion]

very good

and now for x->0

[thorny flame]

miyo

[north lion]

,w \int_0^{\infty}\sin(x)e^{-x}dx

[thorny flame]

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[north lion]

yes

so now we only need to do 0-(-1/2) and we have 1/2

[strong sable]

Tysm!!

It seems that i must be struck in calculation somewhere before

[north lion]

yeah

also credits to the other guy

that helped

[strong sable]

👍 thank you both

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[frail reef]

can someone help me see why the flux across this rectangle (assuming delta x and delta y are very small) equals

[frail reef]

$\qty(\pdv{M}{x} + \pdv{N}{y}) \Delta x \Delta y$

[thorny flame]

eugene_krabs_has_cake

[sweet shard]

do you know the definition of flux?

[frail reef]

$\oint_C \mathbf{F} \cdot \mathbf{n} \dd{s}$ right?

[thorny flame]

eugene_krabs_has_cake

[sweet shard]

that's not the definition, that's a line integral. it's also one side of green's theorem

[frail reef]

well in 2D i know it quantifies the amount to which the vector field flows out of the curve, so it sums up the component of F in the direction of the normal to the curve

but i think in this case i need to split up the closed line integral into 4 line integrals and calculate that, but im not really sure how to

even though i know the normal in each of those cases

[sweet shard]

sorry i was a little wrong. i was thinking flux in terms of a surface

depends on your book's definition, so you should just look that up and start there

[frail reef]

yeah im just caring about the flux across a line integral, instead of across a surface, i think this is for deriving the divergence in 2D

[frail reef]

this is the only real "definition" of flux that i have

[wary tulip]

,av eugene_krabs_has_cake

[thorny flame]

eugene_krabs_has_cake's Avatar

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[sweet shard]

the Flux part at the bottom may help
https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/4%3A_Integration_in_Vector_Fields/4.6%3A_Vector_Fields_and_Line_Integrals%3A_Work%2C_Circulation%2C_and_Flux

[frail reef]

i may be being stupid but i dont understand this last line

[sweet shard]

y(t) is just 1

[frail reef]

for this im not too sure how to make this general with M(x,y) and N(x,y)

[sweet shard]

replace x with M(x,y) and y with N(x,y)

[frail reef]

but if we go from the top line, would r(t) be (x + (delta x) * t, y + delta y)?

or smth close

[frail reef]

but when you evaluate the integral wont you have something like M(x(t),y(t)) and N(x(t),y(t)) which you dont know how to evaluate?

[sweet shard]

yea. just do some problems and see

example 1 and 2
https://tutorial.math.lamar.edu/classes/calciii/surfintvectorfield.aspx

[frail reef]

but im not doing the flux of surfaces, just lines for now

[sweet shard]

oh right i keep forgetting

https://courses.lumenlearning.com/calculus3/chapter/applying-line-integrals/

https://ocw.mit.edu/courses/18-02sc-multivariable-calculus-fall-2010/d4d8e457e8dd0c8c92f24234b52e87ae_MIT18_02SC_MNotes_v3.pdf

https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals-in-vector-fields-articles/a/flux-in-two-dimensions

[frail reef]

if i label c1 as the bottom side, would there be any truth in this equation?

i feel like its wrong since M and N are functions of x and y which would be functions of t aswell so it wouldnt just be a constant in the integral

this is the exact thing im trying to figure out

i just want a bit of help into the explanation of it, preferably the top 2 lines, since the left and right will probably follow easily from that

[sweet shard]

which part of the top two lines don't you follow?

F = M ihat + N jhat

So F dot j = N

[frail reef]

just how they really derived that the flux across the top was .....

[frail reef]

yeah

[sweet shard]

do you know what n is here?

[frail reef]

in the case of the top its j

[sweet shard]

yes

[frail reef]

i understand the equality after this

just not that bit itself

so we have the n being j

[sweet shard]

it's just approximating the integrand as constant

[frail reef]

so im trying to compare it to $\int_{C_1} F \cdot n \dd{s}$

[thorny flame]

eugene_krabs_has_cake

[sweet shard]

since delta y and delta x are small

[frail reef]

so are you saying that they are basically just doing $$\int_{C_1} \mathbf{F} \cdot \mathbf{\hat{n}} \dd{s} \approx \mathbf{F} \cdot \mathbf{\hat{n}} \cdot |C_1|$$

?

[thorny flame]

eugene_krabs_has_cake

[topaz sinew]

@frail reef Has your question been resolved?

[sweet shard]

yes

[frail reef]

so im assuming $\int_{C_1} \dd{s} = |C_1|$ is something thats always true, but i dont understand why we can treat the integrand as a constant since the components of F are functions of x and y

[thorny flame]

eugene_krabs_has_cake

[sweet shard]

you're just linearizing

F(x, y + dy) is approx constant if dy is small enough

[frail reef]

when would F actually be a constant?

in which situation

[sweet shard]

do you know what linearizing is?

[frail reef]

is it $F(x, y +\Delta y) \approx F(x, y) + \pdv{F(x,y)}{y} \cdot \Delta y$?

[thorny flame]

eugene_krabs_has_cake

[sweet shard]

actually simpler, just the first term

[frail reef]

only because delta y goes to 0 right?

or is it just a crude approximation

ok so are we saying that F(x,y) is constant with respect to whatever variable we are integrating?

[sweet shard]

yea the assumption is delta y here is small enough for you to ignore

[frail reef]

i just dont see why we can say F(x,y) is constant, my brain might be fried tbh, this seems like its easy

[sweet shard]

because dx and dy are sufficiently small

it is indeed easy

[frail reef]

but arent we saying F(x,y+delta y) is constant because it can be approximated by F(x,y) and F(x,y) is constant?

is delta x and delta y some kind of variables and x and y are constants here?

[sweet shard]

you're overthinking it

but yes x and y are constant

[frail reef]

Ok so am I thinking about it in a correct way if I say that the only thing that varies in the flux integral is F, and over that line, y stays the same, x goes between a coordinate x and x + delta x

And when we evaluate F on that line we get F(something, y) where that something goes between x and delta x, but wherever that x is evaluated at, between x and delta x, the function is basically the same as F(x,y)

(This is for the bottom line)

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@frail reef Has your question been resolved?

[twin heath]

Crazy username

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[glacial nexus]

Hello! I was hoping to get some help with a Domain/Range problem in my College Algebra (1148) class. There are two straight lines that are perfectly horizontal on the graph. I was assuming it would be R[6]U[8] but wanted to verify if thats the right answer and if not what else it would be?

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[sage jetty]

hallo

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[sharp dew]

I luv you

Let me try that one

[versed cairn]

Lim x approaches to 0 sinx/x is equal to something else

It's a pretty well known limit

Sure

[sharp dew]

I actually stop solving it once justtopro started taking over it

He’s a better man than me

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[ornate beacon]

i get how the first 3 are true but how is the last one converse of corresponding angles

[ornate beacon]

waut nvm i get it

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[ornate beacon]

i js need someone to check my answers
13. x= -5 14. x = 8 15. x = 3 16. x=7

and this is x = 9?

[ornate osprey]

yep looking good

all correct

[ornate beacon]

TYSM

[neon iron]

sorry for disturbing but please remember to close the channel to make it available to others

[ornate beacon]

okay sorry i will

[neon iron]

a is equal to f because of alternate angles

alternate angles are angles that are in between lines that complete a "Z"

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[ornate beacon]

a?

[ornate beacon]

its asking abt f

c and f are alt interior

but what other proofs or steps do i put to get to that

[neon iron]

that i know of

[sonic ibex]

yo what you need help with

[sonic ibex]

ah i see you started proofs

[sonic ibex]

b and c verticle angles and c and f are alternate interior

but also write what is given as the first statment

[ornate beacon]

HI AGAIN

[ornate beacon]

okay i js gave up on that problem ☠️ but thanks

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@ornate beacon Has your question been resolved?

[ornate osprey]

since AC ≅ BC, AD = DB, and both triangles share CD, you can prove congruency through SSS.

[ornate beacon]

alr

thanks

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[somber cairn]

is an undefined slope an inflexion point?

[loud oasis]

no, an inflection point is a point where the second derivative changes signs

[somber cairn]

okay thank you

wait but like it becomes undefined so that doesnt work?

if after the undefined slope it changes sign then does it count?

[loud oasis]

well if the first derivative doesn't exist then the second derivative doesn't either. so it is potentially a place where the second derivative changes

[somber cairn]

oooo okie ill keep that in mind thank u!!!

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[neon iron]

How to make an equation of this?

[agile harness]

what do you think

[neon iron]

I tried to put some numbers but I couldn’t figure out

[agile harness]

so one investment pays 18% meaning youll get an additional 18% of the amount of money you put in that account and same logic for the other account but 12%

let x = the amount invested into the 18% account

let y = the amount invested into the 12% account

what can we say about the amount we have a year later

if it pays 18%

mathematically what does that mean

[neon iron]

The amount is going to increase

[agile harness]

right so we can say x', the amount of money in the 18% account in a year will be 1.18x

right because its x, the initiall amount plus the 18% of the initial amount it pays in a year

so x + 0.18x = 1.18x

[neon iron]

Then what do I do with 12%?

[agile harness]

same logic

using y instead

i dont need to write out the entire solution for you

[neon iron]

y+0.12y=1.12y

Then how do I solve each investments

Like how do I solve just by these two equations

[agile harness]

@neon iron

so what do you know about the sum of 1.18x + 1.12y

[neon iron]

Is it equal to 27480?

[agile harness]

yes

so thats one equation

what other equation can we make

from x + y

[neon iron]

0.18x+0.12y=24000?

[agile harness]

no

x and y are the initial investments

and they sum to 24000

so x + y = 24,000

and 1.18x + 1.12y = 27,480

then you have a system of equations

do you know how to solve from here

@neon iron

[neon iron]

Is 10000 for 18%

And 14000 for 12%

[agile harness]

yes

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[wintry salmon]

.close

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[autumn plaza]

excuse me

[autumn plaza]

how do i do thjis

[fallow heart]

Well, just use $z^{-1} = \frac{1}{z}$

[thorny flame]

Alberto Z.

[autumn plaza]

oh.

..,..

[autumn plaza]

and i need helkp

for this

what is modulus

i dont get it

it doesnt explain what it is

[fallow heart]

It does

[autumn plaza]

wat does it mean

[fallow heart]

That is the definition

[autumn plaza]

????

so its not liek

pythagorean theorem gives u longest side of a triangle

or

[fallow heart]

Ahn you mean the geometrical meaning?

[autumn plaza]

quadratic formulas give u roots of an equation

[autumn plaza]

yess

[fallow heart]

Alright sorry didn't understand

In the end, it's an application of Pythagora's theorem to a triangle having Re(z) and Im(z) as sides

[autumn plaza]

wait

why do we learn imaginary numbers

if theyre fake

[fallow heart]

Because they help a lot in some fields, like physics and engineering

You can use them also to prove trig formulas

[hoary burrow]

Imaginary is a bit of a misnomer

[autumn plaza]

whats a misnomer

[hoary burrow]

A misleading name

Like how the horned toad is actually a species of lizard

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[sacred cloud]

hi im opening this in a new channel because mine closed since i couldnt get back to it in time hhhh

we had a quiz earlier and i js wanna see if my answers are correct, ive never encountered a summation within a summation so i wasnt sure how to answer it but this is what i put:

a) 10r - 14p or 10r - 13p
b) -16 or -18