help-39

[pearl pond]

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[livid tiger]

is the parametric form here correct?

[livid tiger]

i got {4,0,1} + x3{0,0,1}

.close

[pearl pond]

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[pearl pond]

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[regal sequoia]

"The function f at its critical point a has the quadratic form Q(h), so that f(a+h) = f(a)+ 1/2!Q(h).
Determine the signature of the quadratic form Q(h) and determine whether a is a local extremum or not, if..."

[pearl pond]

@regal sequoia Has your question been resolved?

[keen rain]

Does Q(h) refer to Q(h,0) or why is k not mentioned

[regal sequoia]

I honestly have no idea

Maybe he meant Q(h,k)?

[pearl pond]

@regal sequoia Has your question been resolved?

[pearl pond]

@regal sequoia Has your question been resolved?

[rough forge]

I think h is a point/vector but of what components

But the first sentence implies

f(a) = Q(h)

So that means f'(a) = Q'(h) = 0

[regal sequoia]

Hmm yes okay

[rough forge]

Yea sorry

I would interpret h as

(h1, h2)

Q(h) = Q(h1, h2)

[regal sequoia]

Okay

I am not sure what he means by determine the signature of the quadratic form

[rough forge]

,,f(a+h) = f(a) + \frac{1}{2!}Q(h) = Q(h) + \frac{1}{2!}Q(h)

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

I think like determine Q(h)

[regal sequoia]

Oh okay

[rough forge]

I subbed Q(h) = f(a) because of the first sentence

[regal sequoia]

Oh yeah

[rough forge]

We somehow have to involve Q(h,k) so that we figure Q(h)

So maybe

[regal sequoia]

Is the signature not determined by how many positive or negative eigenvalues there are?

[rough forge]

If Q(h) = f(a) then f(a+h) = Q(h+h)

,,Q(2h) = Q(h) + \frac{1}{2!}Q(h)

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

Now plug in for (a)

[regal sequoia]

What do I plug in exactly?

[rough forge]

,,6(2h_1)^2+6(2h_12h_2)+3(2h_2)^2 = Q(h) + \frac{1}{2!}Q(h)

[jolly parrot]

𝔸dωn𝓲²s

[regal sequoia]

Ohh I see

[rough forge]

Maybe solving this for Q(h)

,,Q(\text{h})= \frac{6(2h_1)^2+6(2h_12h_2)+3(2h_2)^2}{ \left(1+ \frac{1}{2!}\right )}

[jolly parrot]

𝔸dωn𝓲²s

[regal sequoia]

Q(h) = 16h12 + 16h1h2 + 8h12?

[rough forge]

This reminds me of something with Eigenvalues but I cant tell what

The quadratix terms can be put intoa matrix

And them from that matrix calculate eigenvalues

[regal sequoia]

Is it not if the leading term is positive it has an extreme point for example a should have a local minimum point and has the signature (++)?

[rough forge]

What I remember lol

Oh that I dont know quite

Did you learn or study somthjng like that

It could be something like that with h1

[regal sequoia]

i mean this is like the only time he has gone throught this

is just says that it has a local extreme point if a is one of these things

[rough forge]

$\begin{pmatrix} 24 & 24 \ 24 & 12\end{pmatrix}$

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

Also the proof task I figured it out with someone yesterday, it's actually easy but I will be gone for some hours, maybe that helps you

[regal sequoia]

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

I got this problem wrong

[midnight haven]

the answers shown are the correct ones

how are you supposed to solve this kind of problem?

trying to get the 2nd derivative, and its a mess

1st derivative

2nd derivative (without simplification)

trying to solve that 2nd derivative mess for 0 I get stuck

[stark summit]

there is a mistake in your 2nd derivative

[midnight haven]

I don't think so

[stark summit]

you missed a x near the 10

it should be -(10x).(10x)

[midnight haven]

yep you are right

my bad

anyways though, how do you solve this for 0?

looks like multiply both sides by (5x^2+7)^2, then simplify the numerator and factor that?

looks like it wont factor

[stark summit]

wdym with solve this for 0, like =0?

[midnight haven]

The question is asking for the intervals of when the original function is concave up or concave down. solving the second derivative for 0 will give me the inflection points, which are the points between the change between being concave up and down

[stark summit]

yeah it works then

[midnight haven]

yeah

fixing the error you told me about is making it go a bit differently

still don't quite get it

[stark summit]

why did you plus 100x?

[midnight haven]

oh

yeah i see something cheeky

[stark summit]

you are forgetting something

[midnight haven]

now what oof

[stark summit]

a minus

50x²-100x² remember

[midnight haven]

ah

[stark summit]

should you solve for 0 tho?

then the equation looks something like (-50x²+70)/(5x²+7)²=0

[midnight haven]

i multiplied both sides by 5x^2+7

oh

then i just get rid of it

[stark summit]

yeah

then you have a simple quadratic

[stark summit]

yes

[stark summit]

yes

[midnight haven]

that looks nothing like the answer

something else is wrong

[stark summit]

the answer is sqrt(35)/5?

then we have the answer

[midnight haven]

yes

[stark summit]

because sqrt(7/5)=sqrt(35)/5

[midnight haven]

OH

nice

thanks so much

[stark summit]

ei no problem

[midnight haven]

I do it right next time

.solved

[pearl pond]

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[pearl pond]

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[dawn sphinx]

For number 1, my answer isn’t matching up with calculators online, what did I get wrong?

[eternal gulch]

what is droping the calculators online @dawn sphinx

[dawn sphinx]

droping?

[eternal gulch]

What result does it give you, I mean

[dawn sphinx]

It says there’s no inflection points and it only conclaves upwards

[eternal gulch]

2 doesnt looks like a inflection point

let me check what its happen

[eternal gulch]

yeah you miss prove that 2 is critical point

If two were the inflection point, then it would have to be the critical point.

[dawn sphinx]

So I solve the root for the first derivative

But then what do I do after that?

[eternal gulch]

You evaluate the points you obtained from solving the first derivative in the second derivative, you look at the sign of the evaluation to know what they are.

[pearl pond]

@dawn sphinx Has your question been resolved?

[pearl pond]

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[plain berry]

How do I find equivalent ratios

[thin bane]

can you give an example? or your exercise or whatever if you ahve any

gives more of a context of your level

[thin bane]

well do you know how multiplication works?

[plain berry]

Ya

[thin bane]

whats 3 times 2

and whats 5x2

[plain berry]

Oh I think I get it now

[dapper bough]

Beezer what grade you in btw

[plain berry]

7

[dapper bough]

Oh yeah that makes sense

[plain berry]

If I didn’t get two pages done my mum wouldn’t let me eat later

So I needed help

[thin bane]

💀

[dapper bough]

☠️

[plain berry]

?

[thin bane]

nothikng @ me if you have any questions

[plain berry]

Ok

[dapper bough]

Syrex

COuld you help me

in help-24

[plucky needle]

Hey

I gotchu

[dapper bough]

Thanks

[pearl pond]

@plain berry Has your question been resolved?

[pearl pond]

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[blazing socket]

Long story short I got index of 4 to 25 inside my radical is there a way I t can be further simplified? Or that’s it? Because from what I know the 25 can’t go in 4 times only 2 times

[tropic saddle]

not sure exactly what you mean

$\sqrt{4^{25}}$ ?

[jolly parrot]

Denascite

[blazing socket]

No I mean this

Can it be further simplified? Or that’s it?

My index is 4 and the inside number is 25

Yes

[jolly parrot]

faiyrose

[blazing socket]

Okay thxxx!!

Okay

Becuase the final answer on the computer was sqrt5

But I didn’t understand how though

[jolly parrot]

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

faiyrose

faiyrose

faiyrose

[blazing socket]

I see so what you did is squared 25 then you multiply the 4rth root 1/4 with 2/1 which gets you 5^1/2

Can you give me another example similar to what you can do that with?

[pearl pond]

@blazing socket Has your question been resolved?

[blazing socket]

Yes thank you

Can you please, explain why that’s allowed because initially I thought it could not be further simplified for future reference.

[pearl pond]

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[pearl pond]

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[rancid sierra]

sorry for text dump. Its a first order logic question with the answer

but i dont understand the answer

[rancid sierra]

So what i dont get it question 9) asks you to prove that F is in consistent.

So my thinking is that i would somehow have to prove that all sub sets of $\Sigma$ are consistent but the answer is not that. The answer just shows one interperation $I'$ where it works

i dont understand how this answers the questions

[jolly parrot]

Br;an

[pearl pond]

@rancid sierra Has your question been resolved?

[rancid sierra]

I will provide a visual aid of my understanding

so the answers says just assumes some largest n exists such that t < w called n is true so re-map I(w) = n + 1 so problem sovled but

unless i missunderstood ${t < w : \text{t is a closed L-Term} }$

[jolly parrot]

Br;an

[rancid sierra]

okay acctually new question:
what does ${t < w : \text{t is a closed L-Term} }$ mean? i assumed this would contain predicates that were false i.e. 3<2 and such forth

[jolly parrot]

Br;an

[rancid sierra]

<@&286206848099549185> ??

[rancid sierra]

perhaps my question is poorly defined

any text book recomendations on first order logic for cs?

alr acc i think i get it

.close

[pearl pond]

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[pearl pond]

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[prime heron]

i cannot for the life of me draw this so i can switch the bounds

[prime heron]

I need to be able to do this without desmos (and i lowkey can't even do it with desmos)

this is what i got so far

I can typically draw these if it's just with two basic regions, like z1 = ... and z2=... but once you add the bounds of the other integrals in there, im done for

[wet osprey]

Why do you have to swap bounds

Does it tell you to

[prime heron]

oh wait sorry this one's not one of the impossible ones, but some of them are

it says swap if its impossible, so would this one just be too complicated

because if i were supposed to swap, i'd be outa luck lol

[wet osprey]

(Btw I also am not too good at switching to spherical cords if that’s what you want to do)

[prime heron]

spherical coordinates shouldnt come into this since this homework isn't based on that lecture, maybe the impossible ones are just simpler

[wet osprey]

They usually are

[prime heron]

ok dude i was obviously panicking too much to realize this is actually a simple one 😭 thanks tho i'll prob be back here soon

.close

[pearl pond]

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[wet osprey]

Aren’t you missing the y = sqrt2 line

[prime heron]

i was confused about if they were planes, lines, etc so i just omitted it lol

[wet osprey]

They are planes

[prime heron]

i didnt even know how to draw it on anything other than my xy plane graph

[wet osprey]

Everything are surfaces

[wet osprey]

It’d just be a horizontal line at y = sqrt2 on the xy plane

It’s also a vertical line at y = sqrt2 on the zy plane

But now you’re just looking at a plane from different angles

[prime heron]

ok ok so for the zx plane it wouldn't show up on that one at all right

yeah i draw three graphs just bc it helps me visualize it

[wet osprey]

Well no because out there page is y

[prime heron]

yeah

[wet osprey]

And y = sqrt 2 is not on the page

[prime heron]

ok got it that makes more sense, idk i need remedial visualization classes lol

thank you!

[pearl pond]

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[crimson echo]

How do I do 13 I know it’s with x and y sides but I need help

[acoustic path]

can you set up two equations to model this situation?

[crimson echo]

No I don’t know how to I been trying to figure it out

[acoustic path]

alright

so forgetting about this situation

do you know how to calculate the volume and surface area of a rectangular prism?

[crimson echo]

A=xy and idk ab volume

[acoustic path]

that's the area of a rectangle

we want to consider a rectangular prism

[crimson echo]

Is it a=x^2y

[acoustic path]

like, one of these

[acoustic path]

this is the volume

do you know what the phrases "Volume" and "Surface Area" mean ?

[crimson echo]

Yes I’m just more familiar with 2d optimization problems

[acoustic path]

i see

that makes sense since normally those are what you're introduced to first

and the surface area is calculated by finding the areas of each of the six rectangles (so, the six faces) of the prism, and adding them together

[crimson echo]

So 6x^2 +8xy?

[acoustic path]

not quite (but you do use the right variables, your answer should be something with x^2 and xy)

keep in mind the "material costs" of 1$ and 2$ per square foot refer to the areas not the lengths

[crimson echo]

So 16=2x^2y

Ignore the picture I started it wrong

How do I set it up because I found y=8/x^2

[indigo pewter]

,rccw

[indigo pewter]

shouldnt SA = 2x^2 + 4x(16/x^2)?

[crimson echo]

I’m confused how do I incorporate the price values in

<@&286206848099549185>

Is 13 right?

<@&286206848099549185>

[rapid meteor]

Hmmn

Which question

[past kernel]

13th one ig

[rapid meteor]

Yes it is

[past kernel]

whats the answer he got?

[rapid meteor]

13

[jolly parrot]

tori

So, the dimensions that minimize the cost are 2 feet for the side of the square base and 4 feet for the height of the box. The cost calculation in the image seems to have an error as it did not follow through with the derivative correctly. The calculated cost with the correct dimensions would be:

\( C = 6x^2 + 96 = 6(2)^2 + 96 = 6(4) + 96 = 24 + 96 = 120 \) dollars

This value represents the minimum cost to build the box given the constraints and costs provided in the problem.

[crimson echo]

So what did I do wrong

[pearl pond]

@crimson echo Has your question been resolved?

[pearl pond]

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[naive dirge]

I suppose it should be correct?

The vertex

I found that g’(x)=0, as well as the vertex, happens at x=-17 / 2

[marble sigil]

the y-value of the vertex is wrong, you'd plug in -17/2

[pearl pond]

@naive dirge Has your question been resolved?

[naive dirge]

Because the vertex happens at x=-17/2

[marble sigil]

yea

it has to be a point on the function right

[naive dirge]

The vertex should be ( -17/2, f(-17/2))

[marble sigil]

yes

[naive dirge]

Yes

f(-17/2)=-9, no?

That’s so weird

[marble sigil]

I get -1/4

it shouldn't dip that low

[pearl pond]

@naive dirge Has your question been resolved?

[pearl pond]

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[glad nimbus]

M^3 - M^2 - 8M + 12

factoring

[steep saddle]

are those M's meant to be different

[glad nimbus]

no

just typo

[vapid mirage]

Cubic equations can be factored by first finding their root by trial and error and then using long division to divide the equation by (m - root). Then you will have (m - root) (quadratic equation) which you can further factor by using the normal method to factorise quadratic equations

In this case if I put M = 2 I get the answer 0. So we will divide the equation by (M - 2)

Which will be (M - 2) (M² + M - 6)

Now we can just factor the quadratic equation further by whatever way we choose

The answer would be (M - 2)(M - 2)(M + 3)

[glad nimbus]

so then just getting m = 2 is guess and check?

[vapid mirage]

Welp, yeaa

That's what I do at least

[glad nimbus]

what if you found m = -3 first

for what ever reason

[vapid mirage]

In that case divide the equation by (x + 3)

Would ultimately result in the same factors

[vapid mirage]

Btw, I just googled and there is something called the rational roots test. You can use that to find the first root. You can google it to see how it works.

The task is to find one root really. You can do it however you like

I have always done the trial error thing myself

[glad nimbus]

the first root being -2 and - 6? so then just dividing using -2?

also sorry im cramming an essay in xd

[vapid mirage]

What do you mean?

[glad nimbus]

sorry i really need to cram this essay

If i get it done within uh 20mins i get extra credit lol

[vapid mirage]

Welp, good luck

[pearl pond]

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[young tree]

in here he's working on the solution

in the video he says f(t) is the integral of sq(t)
and he takes the integral and there's a constant term
so to find the constant term, he finds the a_0 term of the fourier series of f(t) I think, but i don't understand why he uses 1/2pi in front of the integral.

shouldn't the term in front of the integral be 1/half period ?

or is he scaling it as well? I don't understand.

[pearl pond]

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[young tree]

.close

[pearl pond]

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[pearl pond]

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[delicate zinc]

.open

Oh

[fossil jewel]

!status

[pearl pond]

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

[pearl pond]

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[small kayak]

25% and 2 people went to the museum
1/7th of all people went to the library
Others went to the pool

How many people went to the pool if there were 28 people?

[small kayak]

(my brother's homework)

[autumn fossil]

I'd just calculate how many people went to museum, how many people went to the library and then the rest went to the pool

[small kayak]

I mean, 25% of 28 is 7

7 + 2 = 9

Mean while 1/7th of 28 is 4

[autumn fossil]

correct

also correct

[small kayak]

9 + 4 = 13

28 - 13 = 15

[autumn fossil]

yes

and thats the result

[small kayak]

Alright I told him the answer

[autumn fossil]

Try explaining it too

[small kayak]

I did

[autumn fossil]

good

[small kayak]

Another child saved from the evil doom of Percentages

[autumn fossil]

lmao

If that's all, you can close this with .close

[small kayak]

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

If B and A are position vectors of a particle at initial and final point respectively
Kindly draw and show which would be the displacement and how

[midnight haven]

And also tell
How A - B will be displacement
Kindly derive it

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

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[plucky seal]

Heyo

[plucky seal]

,rcw

,lcw

[plucky seal]

The question is find x+y

Tried a few things didnt work

[pearl pond]

@plucky seal Has your question been resolved?

[plucky seal]

A

Dont know where to start

<@&286206848099549185>

[normal frigate]

aa türk müsün

[plucky seal]

Evt

[midnight haven]

geometry is the worst

[plucky seal]

Üçgen geometrisinde bi tık acı çekiyom

[normal frigate]

cevap 255 mi

@plucky seal

[plucky seal]

Evt

[normal frigate]

tmm

bi dk sana atıyom çözümünü

[plucky seal]

Tm

[normal frigate]

y'yi 180 e tamamlayan açıya z dedim

z+y = 180

BF paralel KC demiş

FK uzunluğunu çektiğimde altta küçük bir paralelkenar oldu

FKCB

U kuralından 105 ve 180-z yi yakaladım

z için 2 denklem oldu

105+180-z+x = 360

ve z+y = 180

taraf tarafa topladığında x+y geliyor

z yerine direk 180-y diyebilirsin ama ben böyle yaptım

[plucky seal]

Anladım

Tşk

[normal frigate]

rica ederim

[plucky seal]

Birde bu çizgileri nası çiziyon

[normal frigate]

hangi çizgileri

[plucky seal]

Fk dışındaki her şeye baktım

[normal frigate]

görme işi ya

önce EBDC'yi paralelkenar yaptım ama bir şey çıkmadı

FB paralel KC yi görünce soruyu çözebilirsin

[plucky seal]

İyi soruymuş

Tşk

[normal frigate]

rica ederim

[plucky seal]

.close

[pearl pond]

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[normal frigate]

başka sorun yoksa kapatabilirsin

[plucky seal]

.close

[normal frigate]

kapandı zaten :d

kolay gelsin

[pearl pond]

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[bright bison]

test

[bright bison]

!done

[pearl pond]

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

[bright bison]

.close

[pearl pond]

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[prime heron]

how do you do this?

[prime heron]

I assume you have to switch the bounds because you can't do integral of cos(x^2)

But I don't know how to proceed from there

[warm ledge]

Which bounds are you considering switching?

[prime heron]

I was going to just do dydxdz? I just didn't want dx in the front

[warm ledge]

sure! how would you rewrite this integral to do that

[pearl pond]

@prime heron Has your question been resolved?

[pearl pond]

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[stark bane]

Given quadrilateral ABCD, angle ABC= angle ADC= 90 degrees, BC=2cm, AD=6cm, calculate the area of ​​ABCD

[stark bane]

My teacher suggested that it is a cyclic quadrilateral

[runic shale]

does it come with any image?

[stark bane]

No

That's my problem too

[runic shale]

i suggest you try to draw it

a cyclic quadrilateral like your teacher says

[stark bane]

like this?

[runic shale]

i thought it like this

cyclic

[pearl pond]

@stark bane Has your question been resolved?

[pearl pond]

@stark bane Has your question been resolved?

[runic shale]

nice

[stark bane]

I just realized that the beginning of the article has more information about angle C = 135 degrees

[runic shale]

that helps a ton

we know now that angle A is 45

[runic shale]

i thought about making an external right triangle parting from point C and ending in H

[runic shale]

the angle to the right of point C would be 180-135

(45)

oh wait it doesn’t really help, we don’t know anything else

doest it give the radius of the circle?

[pearl pond]

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[stark bane]

hmmm no

[pearl pond]

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[coral gazelle]

wait just wondering is this from further maths gcse

[kind briar]

0607 igcse

[jolly parrot]

faiyrose

[kind briar]

ya

[coral gazelle]

this is mainly a rationalisation problem

[kind briar]

h is what we have to find

[coral gazelle]

yep

so we have b and the area

[kind briar]

b is 6+2 sqrt7 + x?

area we are given

[coral gazelle]

where does + x come from

[kind briar]

x is the other side

its given bc

OH

IM SO STUPID

WTF

I THOUGHT SMTH ELSE
SORRY

8^x=\frac{\left(2^{56}-4^{26}\right)}{30}

can u help w this tho

how do i do the bot thingy

8^x=$\frac{\left(2^{56}-4^{26}\right)}{30}

8^x=$\frac{\left(2^{56}-4^{26}\right)}{30}

o

$8^x=$\frac{\left(2^{56}-4^{26}\right)}{30}$

[jolly parrot]

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

[kind briar]

ya

$8^x=(2^56 - 4^26)/30

$8^x=(2^56 - 4^26)/30$

no

its 2^56

and 4^26

[jolly parrot]

faiyrose

[coral gazelle]

right

so in the numerator

try getting the same base

[kind briar]

2^56

and 2^52

so that leaves

2^4

by 30?

[coral gazelle]

no no no

you cant do that with subtraction

that's if you're dividing

[kind briar]

ohhhh

i forgot

💀

[coral gazelle]

try factoring something out of the numerator

[kind briar]

wym

like its

2^52 * 2 * 2 *2 times 2

thats 2^56?

idk

[coral gazelle]

what's the common factor

in 2^56 and 2^52

[kind briar]

2?

@coral gazelle

[coral gazelle]

oh mb sry

2 isn't the highest common factor

[pearl pond]

@kind briar Has your question been resolved?

[pearl pond]

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[cloud bear]

What is the motivation for splitting the matrix at the top, and how would I know what to split it to?

[pearl pond]

@cloud bear Has your question been resolved?

[cloud bear]

<@&286206848099549185>

[dire parrot]

@cloud bear I guess since psi is that exp(A(x-x0)) which can be rewrittzn as exp(Ax)exp(-Ax0)

[pearl pond]

@cloud bear Has your question been resolved?

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[dark ingot]

How to find v_c

[rigid mist]

are you gonna give us some other info?

[dark ingot]

This is the only info

A moves with v0 in -x direction
B mobes with 2vo in x direction

[pearl pond]

@dark ingot Has your question been resolved?

[pearl pond]

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[crisp brook]

For this problem couldnt i just draw a 45-45-90 triangle and do sohcahtoa?

[crisp brook]

Or whats another way to do it cos that doesnt work 😭

[quiet halo]

okay so

225 degrees

do you know the unit circle?

[crisp brook]

Ya

[quiet halo]

now if we think about it

[crisp brook]

5pi/4?

[quiet halo]

225 degrees is basically just 180 degrees + 45 degrees

[crisp brook]

So would a 45-45-90 triangle work?

Cos i thought thats how i learned it in class but it didnt work when i tried it

[quiet halo]

which mans that for sin of this is the same as sin of 180 degrees - 45 degrees

so isn 225 degrees = sin 135 degrees

[crisp brook]

Wait huh

[quiet halo]

sin 135 degrees should be a thing written in your formula book that you get

[merry carbon]

(And it can, but then also remember you need to take into account quadrant info for signs)

[crisp brook]

Um 😭

[quiet halo]

okay so

okay

[crisp brook]

OMG HI AGAIN

[crisp brook]

OHHHH

[quiet halo]

yes thats a big thing chart too

[crisp brook]

😭

[quiet halo]

do you know how

if you have sin

you get 2 answers

in the unit circle

[crisp brook]

Wait what

How

[quiet halo]

lemme send a picture

this is 225 degrees

in the unit circle

what does sin corrolate with?

what does sin GIVE here

[crisp brook]

Isnt 225 in the 3rd quadrant

[quiet halo]

wiat

shit

my bad

ehrm wrong number

[crisp brook]

LMAO DW

[quiet halo]

!

right

now

what does sin give here

[crisp brook]

Uhhh

Wait would it be

-sqrt(2)/2

[quiet halo]

it would

but thats not what im trying to aim at here

[crisp brook]

OMG

OH

[quiet halo]

so

the sin value

corresponds with the Y value

the cosine value

corresponds with the X value

[crisp brook]

MHM

[quiet halo]

the lines on the side represent the X and Y value

which the Sin / cos show

now if you think about it

is there any other point in the unit circle where it has the SAME Y value

aka sine value

[crisp brook]

Pi/4?

[quiet halo]

nope

[crisp brook]

Oh 😭

[quiet halo]

bingo!

that side also has the same Y value doesent it?

[crisp brook]

OH

YEAH

[quiet halo]

which means here

sin 225 degrees

which is

sin 180 degrees + 45 degrees

is also

sin 360 degrees - 45 degrees

[crisp brook]

OHHH

[quiet halo]

aka sin 315 degrees

completely not related to the question but still something i want you to know about the unit circle

the same thing can be said with the X value

aka COSINE value

where on the circle would it have the same X value

at 225 degrees

[crisp brook]

Quadrant four?

[quiet halo]

bingo!

[crisp brook]

Hooray!

[quiet halo]

bad at drawing

[crisp brook]

:DD

[quiet halo]

but yeah

[crisp brook]

LOL DW

[quiet halo]

you know what that also means

[crisp brook]

?

[quiet halo]

sin and cosine

HAS 2 ANSWERS

!!!!

[crisp brook]

OHHH

OMG

[quiet halo]

SO

if i say

[crisp brook]

WAIT THATS SO COOL

[quiet halo]

sin x = 3/4

and you do x = sin^-1(3/4)

on a calculator

it only gives you one answer

which my calculator says is

48.6 degrees

BUT

[crisp brook]

Wait why the -1

[quiet halo]

thats arcsine

the inverse of sine

its just to figure out the angle with a calculator

[crisp brook]

OH

[quiet halo]

like what angle gives the y value 3/4

and the claculator says

48.6 degrees

so sin(48.6degrees) is 3/4

WELL

we have 1 more solution

[crisp brook]

180-48.6?

[quiet halo]

close!

think about it

48.6 is in 1st quadrant

whichj means

y is positive

what other quadrant is y positive

[crisp brook]

2?

[quiet halo]

4

[crisp brook]

Huh

[quiet halo]

in 2 y is negative

wait

is my quadrants wrong

its been a while

FUCK

[crisp brook]

2 1
3 4

[quiet halo]

MY QUADRANTS ARE WRONG

YES

2

2

2

2

[crisp brook]

HELPPPPPP

💀💀💀💀💀💀💀

LMFAO

[quiet halo]

hey its been 2 years since i did this ok

i dont remember the names

[crisp brook]

LMFOAOAOAOA

[quiet halo]

i just know it in my head

okay so

yes

quadrant 2

wait actually shoot me please

it is 180 - 48.6!

i was thinking of wrong angle for a second

[crisp brook]

RAHHHH OK 🔫🔫🦅🦅🇺🇸🇺🇸

[quiet halo]

but THE POINT STANDS

you understand me right

[crisp brook]

YES

[quiet halo]

how it gives you 2 solutions

[crisp brook]

I GET IT

[quiet halo]

WELL

WHAT IF I SAY

THERES AN INFINITE

SOLUTIONS

MUHAHAHAH

[crisp brook]

WHAT.

[quiet halo]

cuz if you think about it

what happens

[crisp brook]

😨

[quiet halo]

if you get to 48.6 degrees

BUT

you decide

to do another round?

and go to 360 + 48.6 degrees?

wellll, you would just get back to the same place right?

[crisp brook]

Omg 😮

[quiet halo]

so

the solution would be

[crisp brook]

And then add 2pi?

[quiet halo]

48.6 + n * 360 degrees

you can add 2 pi aswell

but thats radians

[crisp brook]

Woah

[quiet halo]

where n is a whole number

integer

and 2. 180 - 48.6 + n * 360degrees

[crisp brook]

Wait thats actually so cool

[quiet halo]

the unit circle is a monster

its cool to know it

[crisp brook]

If i get a question like that on a test or something im just gonna cram in as many of these solutions as possible and see if my teacher will take it

[quiet halo]

no just say

the angle

• n * 360 degrees

where n is a integer

[crisp brook]

Do i write this part

[quiet halo]

yeus

cuz

you cajnt do 1.1 times 360 degrees

you need wole laps

[crisp brook]

Ohhh ok

[quiet halo]

not half a lap

[crisp brook]

Okk tysm :DD

[quiet halo]

no worries

[crisp brook]

.close

[pearl pond]

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[pearl pond]

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[crisp brook]

Hi i dont get why this is wrong 😭

[crisp brook]

The question wants two answers

[stark barn]

answers usually between pi/2 and -pi/2

[vale sentinel]

-30 should be 330

[crisp brook]

Wait why

[vale sentinel]

uhh

idk how to explain

-30 and 330 are the same

[crisp brook]

😭

Wait but

If theyre the same

Why dont they both work

[vale sentinel]

uh

[crisp brook]

Does the answer hafta be positive?

[vale sentinel]

Yes.

[stark barn]

-30 and 180-30
-30 and 150

[crisp brook]

OH

Wait does the answer in radians also hafta be positive

[vale sentinel]

idk didnt learn radians yet

[crisp brook]

Oh 😭

[stark barn]

Principal Argument

[crisp brook]

Wait how do u know unit circle but not radians

[stark barn]

should be in the range pi/2 to -pi/2

[vale sentinel]

we only learned degrees in gr 11

[stark barn]

-150 should be 150

[merry carbon]

Do you have the full question though?

[vale sentinel]

i watched some vids on radians but yk didnt get taught yet so idk 100%

[crisp brook]

But doesnt the answer hafta he positive?

[crisp brook]

Ya

[stark barn]

no

[merry carbon]

Can you show the whole context please

[stark barn]

you can add any value of 2pi to the angle and itll work

[crisp brook]

Ahh i see

[crisp brook]

Wait but if u add 360 to it wouldnt it be 210

[vale sentinel]

yes

[crisp brook]

Then for the radians

[crisp brook]

Do i add 2pi to it

[merry carbon]

Those are why it needs to be positive btw, because they tell you

[crisp brook]

OH

OHHHHH

💀

[stark barn]

well that solves that

lol

[merry carbon]

(If they didn’t then both would work)

[crisp brook]

So without the context of this problem negative answers could be accepted?

AKFJAK THANK YOU GUYS :DDD

[merry carbon]

Also an important reason why you should include all context for a question

[crisp brook]

Yeah 😭

I need to start reading the ENTIRE question from now on 💀

[stark barn]

brother lmao

[crisp brook]

STOP 😭😭😭

HOWD U FIND ME

💀

[stark barn]

💀

[crisp brook]

Ok from now on ill ACTUALLY start reading it METICULOUSLY

[coral gazelle]

clip that for future use

[stark barn]

mmHmm definitely

[crisp brook]

😭😭😭😭😭😭

[stark barn]

ill see you in 10mins

in a new chat

[crisp brook]

Noooo 😭

WE’LL SEE ABOUT THAT

Oh oops i forgot to close it

ANYWAYS THANKS GUYS

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

Hey

[midnight haven]

I was messing around with finding areas behind some convergent functions and I got this

You can see that the final area is 2,246. The goal was to get a good estimate value without using integration

I also integrated this function in this interval and got 2 as a result

So, does that mean that my previous estimate result was right? The difference is only 0.2

[midnight haven]

So, is this right?

Keep in mind that I only used 4 rectangles, when I could’ve used more to get a better approximation

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[midnight haven]

I need help

[coral gazelle]

think of a potential u sub

[midnight haven]

this is what I did.

i said u = lnx

du = dx/x

that replaces the top part. so we end up with du/u

[coral gazelle]

mhm

[midnight haven]

integral of du/u = ln(u). so we get ln(ln(x))

is that wrong?

[coral gazelle]

nope

plug in the bounds of integration

[midnight haven]

so u(e^2) = 2 and u(e) = 1

[coral gazelle]

no no no

now you went wrong

you either change the bounds of integration and STAY in the u world
or you keep the bounds of integration and go back to the x world

[midnight haven]

Ohhhhh

so if I were to stay in the u word we would only have ln(2)-ln(1)

which is just ln(2)

[coral gazelle]

yep

and since you went back to the x world

keep the bounds the same

and you get same answer

[midnight haven]

@coral gazelle thank you bro

I appreciate it

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

answer is B right?

.close

[pearl pond]

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[pearl pond]

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[grand bough]

$\tan\left(x-\frac{\pi}{4}+\frac{\pi}{4}\right)$

[jolly parrot]

RizzlerFly

[grand bough]

this should be

$\frac{\tan\left(x-\frac{\pi}{4}\right)\ +\ 1}{1-\tan\left(x-\frac{\pi}{4}\right)}$

[jolly parrot]

RizzlerFly

[grand bough]

but look how they wrote it

[pearl pond]

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[grand bough]

noo

[pearl pond]

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[nocturne patio]

Im reviewing probability for an exam, how is the answer here 155?

[pearl pond]

@nocturne patio Has your question been resolved?

[final hinge]

The average (expected) times you get 5 heads for each trial is 1/2^5

So you'd need on average 2^5 trials to get 5 heads

Which is 160 flips, and -5 to get the previous number of flips

[pearl pond]

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[tiny trout]

The radius of a right circular cylinder is given by √t+2 and its height is 1/2√t where t is time in seconds and the dimensions are in inches. Find the equation for the rate of change of the volume with respect to time.

[tiny trout]

This is an implicit differentiation question right?

I'm fairly sure this is corrected, but my friends seem to think normal differentiation would suffice. However, I believe that since this is a related rates problem, implicit differentiation is required.

[pearl pond]

@tiny trout Has your question been resolved?

[woven matrix]

the pb seems to come from product rule

why didn't the product of r and h give 2 terms after derivation

[final hinge]

You want $\dv{V}{t}$, so either you write $V(t)$ (in terms of $t$), or use the prof rule like
$$\dv{V}{t} = \dv{(r\cdot h)}{t} = h\dv{r}{t} + r\dv{h}{t}$$

[tiny trout]

oh no. I did differentiate wrong

[jolly parrot]

π=√g

[tiny trout]

I got up to this, but I don't know how to simplify it further

[woven matrix]

you can put the 2 fractions on the same denominator

note that if like your friend, you did direct differentiation, you should get the same result

[tiny trout]

Should I just do direct differentiation? Because that sounds much more straight forward.

Also, why can either work? I thought because it was related rates that we could only use implicit

[woven matrix]

it all depends only on t
so when you write V = pir²h, you mean V(t) = pir²(t)*h(t), where V, r and h are function of t
then, when you want to know dV/dt, it's simple differentiation with respect to t:
pi*(dr²/dt h + dh/dt r²), whether you replace them by their expression now or later doesn't change anything of that

you can see differentiation like an operation
when V(t) = pir²(t)*h(t), whether you apply it on the actual V(t) (which your friend did) or pir²(t)*h(t) implicitly (which you did) it's obv the same

[tiny trout]

So I think in this situation, explicit is the shorter method

[woven matrix]

well they should be the exact same length

if we exclude the line where you're essentially writing product formula

[tiny trout]

I'm going to close the channel until I figure out the solution

It usually takes me a while

thank you for the help.

.close

[pearl pond]

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[tiny trout]

Is the correct answer now? I decided not to continue with implicit differentiation because I'm not as confident and I'm very short on time.

I have it saved though

.reopen

[pearl pond]

✅

[finite jolt]

Wym very short on time

[tiny trout]

I have a bunch of other assignments, and it's getting late. I need to do another assignment and study for a mock exam, but I've spent all day on calc problem sets. But the context of the prolbem is above.

[tiny trout]

this problem

[pearl pond]

@tiny trout Has your question been resolved?

[pearl pond]

@tiny trout Has your question been resolved?

[tiny trout]

i give up

.close

[pearl pond]

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[tiny trout]

.close

[pearl pond]

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[clear forum]

I am confused on question 1 of the rational model

[clear forum]

How to I make the equation to graph? I thought it was y = 5(2x) but that doesn’t seem right

And when question 2 mentions asymptotes I thought it could be a (x + •••)/(x + •••) equation

I’m just really lost and don’t know how to take my first step

[clear forum]

5(2x) because there are two Xs and fencing per foot is $5

<@&286206848099549185>

: (

[pearl pond]

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[pearl pond]

@clear forum Has your question been resolved?

[pearl pond]

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[hot lodge]

In problem 3 (a), I do not understand what they are asking to show

[pearl pond]

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[pearl pond]

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[winged haven]

what's the point of a slope field? why should we ever figure out one?

[sterile tusk]

a slope field generally shows us what the slope of our function should look like at any given point

it allows us to graph any solution to our function essentially

by just simply connecting them

[pearl pond]

@winged haven Has your question been resolved?

[winged haven]

But then, why not just graph the function to see what it is looks like? Why would there be a need to get an abstraction of the function?

[sterile tusk]

cuz sometimes it might be hard to antidifferentiate and solve the differential equation

which is why we can then use a slope field to see all the possible solutions to that differential equation

[pearl pond]

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[compact ridge]

What do you get for the summation from k = 1 to k = j?

It's implied that you have 1 in the summation of all 3

[chrome flame]

(j-1)?

[compact ridge]

Yeah

So the strategy is to break the sum up so that you have sum of a constant, sum of i, and the sum of i^2

As you move from inner sum to outer sum

[chrome flame]

Or it will be
(i-1)+1-i)

2i

[compact ridge]

Yep so the sum of j from j = 1 to j = i

Will be just $\frac{i(i + 1)}{2}$

[jolly parrot]

south