help-19

[desert marlin]

try usub

[dense patrol]

the numerator doesn't cancel out tho

[desert marlin]

make it cancel

[dense patrol]

lmao yea got it

thank you

.close

[odd edge]

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[wanton escarp]

@thorn palm

[thorn palm]

any trials?

[wanton escarp]

Trials?

[thorn palm]

What have you tried I mean

[wanton escarp]

All that

[thorn palm]

ok

[wanton escarp]

But am I approaching it right

Like finding length of the triangle

[thorn palm]

I solved with length of triangle

so yes you can do it with length

[wanton escarp]

Wdym have u done this question before

[thorn palm]

I solved with at your first ask

the channel you closed

[wanton escarp]

Sorry

[wanton escarp]

Yeah I got that

[thorn palm]

Idk how much help you want

[wanton escarp]

It says on left

Oh so

Is the

Pythagorea one right

The ah

[thorn palm]

the idea itself of using pythagoras?

[wanton escarp]

Yeah

[thorn palm]

or especially the one you wrote?

[thorn palm]

I didn't used the height in my proof

so I'm not sure about that

[wanton escarp]

Ok then I guess not

So then I’m stuck

What do I have to do next

[thorn palm]

Basically this proof is from shkwing that distinct two side has same length

[wanton escarp]

Which 2 sides

[thorn palm]

I used this two

[wanton escarp]

Yeah so that one

I tried using Pythagoras

To solve for that

[thorn palm]

can you write down the length of left orange one?

with pythagoras

[wanton escarp]

Is that a rhetorical question or a real question

[thorn palm]

part of proof

[wanton escarp]

Like do u want me to do it

[thorn palm]

asking to you

yes

[wanton escarp]

Or are u asking if it’s possible

Ok

1st step?

[wanton escarp]

Is that triangles angle same as the other triangle ?

[thorn palm]

do you know coordinate of those two point?

[thorn palm]

yes

[wanton escarp]

Not yet

But it’s the same as other triangle

The same method I mean

[thorn palm]

you know the angle

[wanton escarp]

To find coordinates

[thorn palm]

Do you know that coord of point A=(cosA,sinA)?

[wanton escarp]

Isn’t A an angle

[thorn palm]

sorry

red point

[wanton escarp]

Not yet

Wait what

I’m still getting the red pint as

For x coordinate as

-1 -cosA-B

[thorn palm]

how so?

[wanton escarp]

Wait is it not -1

It has to be be grater than -1

Cause it’s inside of it

[thorn palm]

you mean this?

[wanton escarp]

Yeah

[thorn palm]

that is not the x coord

[wanton escarp]

But that’s the other side

Yeah realised

Is that the other triangle

[thorn palm]

The coord you wrote down here

[thorn palm]

Can you tell me how you got it?

[wanton escarp]

X is aa y is Ao

[thorn palm]

you can do similar thing tk get coord of red&blue point

[wanton escarp]

Very true I know

[wanton escarp]

Ooo

Good info

I didn’t realise that

[thorn palm]

So you can just get the position of red&blue point

[wanton escarp]

To find x coordinate for red

Do I have to find the actual triangles lengths first ?

[thorn palm]

you can do with length 1 sides

[wanton escarp]

Is it one

The triangle across x axis

Can’t be

[thorn palm]

I think you dont have to think about that triangle

[wanton escarp]

But how can u know any lengths of the small triangle

[wanton escarp]

So red was 1

That makes it a lot easier

[thorn palm]

hypo is 1

[wanton escarp]

.

[thorn palm]

Ah sorry I misunderstood your q

yes 1 because circle's radius is 1

[wanton escarp]

That means

(ba,bo)

Is red coordinate

[thorn palm]

not actually

[wanton escarp]

😭

[thorn palm]

aa was coord itself

because aa was on positive side of x axis

[wanton escarp]

So is it just negative x

-bo

[thorn palm]

yes

-ba

[wanton escarp]

Or whichever

He’s

Yea

[thorn palm]

(-ba, bo)

yes

[wanton escarp]

Yea

[thorn palm]

And you can do same thing with blue point

[wanton escarp]

Simple mistake for such a complex question

For blue do u have to use the triangle

[thorn palm]

same method with getting the red point

so if you mean drawing new triangle, yes

[wanton escarp]

How do u find that angle

[thorn palm]

which angle?

[thorn palm]

see what I did for red point

[thorn palm]

here

[wanton escarp]

That plus C

?

[thorn palm]

no

[wanton escarp]

What

It has to be

[wanton escarp]

Yup

That angle where A is but not actually the angle A

Cause it’s 180-A plus that A part

Which is C

[thorn palm]

you can actually just express that angle with only using B

[wanton escarp]

Oh so just 180-B

[thorn palm]

yes

[wanton escarp]

Bruh

[thorn palm]

yep

[wanton escarp]

So now do we prove

[thorn palm]

before that

Do you know sin(180-A)=sinA?

[wanton escarp]

Visually?

[thorn palm]

I mean the fact itself

[wanton escarp]

Now I know

[thorn palm]

do you know that fact

So can you get that red point is (cosA,sinA)?

[wanton escarp]

Where does the minus go

[thorn palm]

cos(180-A)=-cosA

[wanton escarp]

K

[thorn palm]

Now you can prove it

using that two orange line's length is same

[thorn palm]

.

[wanton escarp]

………

How do I prove it ?

[thorn palm]

Can you express each orange line's length?

with pythagoras

asking to you

[wanton escarp]

I did for one

[thorn palm]

you mean $a_h$?

[clever fjord]

Dri111

[wanton escarp]

Yeah

[thorn palm]

that is neither orange segment

[wanton escarp]

Uhh yes it is

[thorn palm]

Im so sorry

I confused it as height

[wanton escarp]

It’s the hypo

[thorn palm]

so you know right orange one

what about left orange one?

[wanton escarp]

Do I just make ah=

[thorn palm]

nope

with red&blue coord

btw blue coord becomes (cosB,sinB)

[wanton escarp]

Am I suppose to find another tangle

[thorn palm]

no we found coord of red & blue point right?

[wanton escarp]

Yeah

[thorn palm]

and left orange is line connecting red and blue

[wanton escarp]

Yeah

So minus does

Those

?

[thorn palm]

Have you learn how to calc distance of two point before?

[wanton escarp]

Idk

Depends

How it works

But I can’t think of a way to rn

So probably not

[thorn palm]

https://images.app.goo.gl/zj5XQJz65Z8WFHQk6

I think this image depicts well

[wanton escarp]

Ok yeah

[thorn palm]

so you can calc length with red&blue coord

[wanton escarp]

B4 that

How do u calculate

The other orange line

[thorn palm]

right one?

or left?

[wanton escarp]

That formula

This equation

[thorn palm]

we will just let it be for now

for sake of simple calculation

[wanton escarp]

@thorn palm

[thorn palm]

yes

so now we use d=ah

[wanton escarp]

How tf do we simplify that

[thorn palm]

square both side first

to get rid of square root

then expand all squares

[wanton escarp]

How do u expand cos b -cos a)^2

[thorn palm]

$(a-b)^2=a^2-2ab+b^2$

[clever fjord]

Dri111

[thorn palm]

you can use this formula

[wanton escarp]

💀

[thorn palm]

you have to do that 🙂

what did you got for RHS?

[wanton escarp]

Idk how to expand

The

[thorn palm]

Actually

I think you wrote it wrong for ah

[wanton escarp]

Well I haven’t starting expanding yet

[thorn palm]

$\sqrt{(1-\cos{(A-B)})^2+(\sin{(A-B)})^2}$

[clever fjord]

Dri111

[thorn palm]

you need to put 1 inside square

[wanton escarp]

Ok

And do I still expand

[thorn palm]

yes

[wanton escarp]

Do u have the expanded version on u

[thorn palm]

I did expanded it too

but idh the writing rn

[wanton escarp]

Is that right

[thorn palm]

Is middle one you wrote 2cos(A+B)?

[wanton escarp]

Yes

[thorn palm]

why?

[wanton escarp]

I don’t know how to expand that one ngl

[thorn palm]

same $(a-b)^2=a^2-2ab+b^2$

[clever fjord]

Dri111

[thorn palm]

in this case a=1 and b=cos(A-B)

btw I recommend writing ${\cos{(A-B)}}^2$ as $\cos^2(A-B)$ not $\cos(A-B)^2$

[clever fjord]

Dri111

[thorn palm]

yes

so you can write as 1-2cos(A-B)+{cos(A-B)}^2 simply

[wanton escarp]

Now the sin part

[thorn palm]

$(\sin(A-B))^2$ right?

[clever fjord]

Dri111

[wanton escarp]

Yeah

Is that expanded already

[thorn palm]

nothing to expand really

just $\sin^2(A-B)$

[clever fjord]

Dri111

[wanton escarp]

Ok now we start eliminating ??

[thorn palm]

yes

[wanton escarp]

Is sinA+B)^2the same as sinA^2 and sinB^2

[wanton escarp]

Or

[thorn palm]

do you think $(a+b)^2=a^2+b^2$?

[clever fjord]

Dri111

[wanton escarp]

No

[thorn palm]

so yeah, not the same

[wanton escarp]

How on earth do I simplify this

[thorn palm]

and where does sin(A+B)^2 comes from?

[wanton escarp]

Meant to be a-b

[thorn palm]

do you know the fact that for real number x $\sin^2 x + \cos^2 x =1$?

[clever fjord]

Dri111

[wanton escarp]

Now I know

[thorn palm]

is this question from lecture or class?

[wanton escarp]

Hs class

Like a practise problem

[thorn palm]

what are you learning about rn?

[wanton escarp]

Trig

Well this book is like scholarship questions which are extended

And we still are learning trig calc normal version

Simply this book has stuff we haven’t learnt

[thorn palm]

And it is first time you saw this?

[wanton escarp]

Yeah

[thorn palm]

Interestinf

ok

so you can make some squared trigs into 1

[wanton escarp]

Is that only on rhs

Cause x has to be in front of sin a^2?

[thorn palm]

both side

[thorn palm]

wdym?

[wanton escarp]

Idk but I found these

If not can u work ur magic and simplify for me

@thorn palm ?

[thorn palm]

yes you can combine those

so $\cos^2B+\sin^2B=1$

[clever fjord]

Dri111

[thorn palm]

sorry my Internet is not properly working rn

[wanton escarp]

So we have 4 on lhs

Then 2 on rhs ?

From

[thorn palm]

yes

4 on lhs?

(cosB)^2+(sinB)^2=1

[wanton escarp]

2 I mean

[thorn palm]

ok

then continue eliminating

[wanton escarp]

Is it 2 both sides?

[thorn palm]

yes

[wanton escarp]

Do I get rid of

Both those

[thorn palm]

they become 1

[wanton escarp]

Does that whole 2 things go away

[thorn palm]

so yes if I understoood your question correctly

[thorn palm]

if you mean (cos(A-B))^2 and (sin(A-B))^2 both disappears, then yes

[wanton escarp]

Yes

[thorn palm]

didnt you say 2 on lhs?

[wanton escarp]

Yea mb

[thorn palm]

and cos(A-B) on rhs

and where did 1 go on rhs?

[wanton escarp]

The 1-2?

[thorn palm]

lhs should have 2

and rhs also should have 2

[wanton escarp]

Ok

Like this?

[thorn palm]

RHS was $1-2\cos(A-B)+\cos^2(A-B)+\sin^2(A-B)$ right?

[clever fjord]

Dri111

[wanton escarp]

Yeah

[thorn palm]

and $\cos^2(A-B)+\sin^2(A-B)=1$

[clever fjord]

Dri111

[wanton escarp]

Yup

[thorn palm]

So $1-2\cos(A-B)+\cos^2(A-B)+\sin^2(A-B)=1-2\cos(A-B)+1$

[clever fjord]

Dri111

[wanton escarp]

Yes

[thorn palm]

therefore RHS is 2-2cos(A-B)

[wanton escarp]

Does the 2-2

Go to 0

[thorn palm]

right 2 is multiplied with cos(A-B)

[wanton escarp]

Ok

[thorn palm]

same as 2-2×3 is not 0

[wanton escarp]

Yes

[thorn palm]

then now you can eliminate more

[wanton escarp]

Could u finish it off please

[thorn palm]

you should do it yourself

[wanton escarp]

I’ll try than

The 2cos(a-b)

Cancels out?

[thorn palm]

wdym

are there 2cos(a-b) on lhs?

[wanton escarp]

Yeah

[thorn palm]

where?

[thorn palm]

that one is just a steps for showing that rhs should be 2-2cos(A-B)

[thorn palm]

you should correct this equation eith that

[wanton escarp]

The final equation has to be

[thorn palm]

yes

[thorn palm]

you should fix rhs with 2-2cos(a-b)

[thorn palm]

because of this

[thorn palm]

you forgot to write 2 for sinBsinA at LHS

[wanton escarp]

The 2 is missing

Yeah

[thorn palm]

now eliminate

[wanton escarp]

?????

[thorn palm]

how did upper line become next one?

[wanton escarp]

I just brought the 2-2 to the rhs

[thorn palm]

2cosBcosA is multiplied together

they should move together

on LHS

[wanton escarp]

So

Bring the whole thing

To the rhs?

[thorn palm]

just eliminate 2 on both side

in other word substract 2 on both side

[wanton escarp]

So then all the 2s are gone?

Or the

[thorn palm]

yes

[thorn palm]

divide both side by -2

[wanton escarp]

This right here?

[thorn palm]

for this just divide by 2

[wanton escarp]

And the divide by 2 on all sides?

[thorn palm]

yes

[wanton escarp]

W w

In the chat

Holy

How long did that question take u

[thorn palm]

about 3~4min

[wanton escarp]

I understand everything but just the simplifying parts I’m iffy about

[thorn palm]

dont worry

pratice would make you better

[wanton escarp]

Well thanks for this super long journey

[thorn palm]

np

btw

[wanton escarp]

Thanks for ur time

[thorn palm]

i dont accept firend req

pls read my profile

[wanton escarp]

Ik I just did cause of ur bio

But thanks 🙏

[thorn palm]

have a nice day

[wanton escarp]

🫡

U too

.close

[odd edge]

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[sonic sail]

This is what i did so far, how to continue further?

<@&286206848099549185>

[fluid tundra]

||substitution|| ||u = pi - x||

[sonic sail]

It will expand even more?
Since there would a square of (pi+u)

[steady tide]

not gonna lie this is some integration bee level integrals

[fluid tundra]

true

[fluid tundra]

you will see why this eventually works out

whenever you have some weird trig integral with bounds 0 to pi or 0 to pi/2.. something like that

the solution almost always involve a substitution of that type, and this case is not an exception

in fact the denominator (sinx)^4 + (cosx)^4 at the start may tip you off at that

[steady tide]

^ this is why this isnt an indefinite integrals

or else

[fluid tundra]

lmao

[sonic sail]

I am stuck at the same thing now but with more terms to solve

[sonic sail]

Yea there must be some sort of trick going around here, which i obviously can't catch

[fluid tundra]

what do you get after substitution?

[sonic sail]

3 terms

[fluid tundra]

that is not very specific of an answer 🤔

maybe show / send it

[fluid tundra]

hm nice

[fluid tundra]

so the sum of these three integrals equals the value of the original integral

[fluid tundra]

do you notice anything similar / different about the new integrals you have and the one you have at the bottom of this page?

[steady tide]

the next step is to scour through https://en.m.wikipedia.org/wiki/List_of_definite_integrals and pray the answer is there

[fluid tundra]

entertaining entry lol

[sonic sail]

🥲😂

[sonic sail]

the third term is quite similar to the original ques

[fluid tundra]

it is

then, what happens when you add the two integrals together?

[sonic sail]

they get cancelled

[fluid tundra]

right

[sonic sail]

still 2 terms left...

[fluid tundra]

with the cancellation in mind, can you show me what integral we have to evaluate exactly to get our answer?

[odd edge]

@sonic sail Has your question been resolved?

Channel closed

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[odd edge]

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

can someone tell me about steps involved in following factorisation?

[dusty badger]

thats the steps right

ur confused abt the steps

[formal jasper]

yes

[dusty badger]

u want me to explain it here

or want a pic

[formal jasper]

anything works

[dusty badger]

why did I get x-2^2

[formal jasper]

thats my problem aswell

[dusty badger]

wait u got that as well?

[formal jasper]

i tried many different ways but i keep on getting something in additon and subtraction

[dusty badger]

Then the pic must be wrong

use the ,w thing

I think it should help

,w x^3+x^2-16x+20

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ___snowflake

[dusty badger]

@formal jasper

that answer is wrong probably

x³ + x² - 16x + 20

= x³ - 2x² + 3x² - 6x - 10x + 20

= x²(x - 2) + 3x(x - 2) - 10(x - 2)

= (x² + 3x - 10)(x - 2)

= (x² - 2x + 5x - 10)(x - 2)

= [x(x - 2)+5(x - 2)](x - 2)

= [(x - 2)(x+5)](x - 2)

= (x - 2)(x + 5)(x - 2).

Thats it

[formal jasper]

i just saw the question has a negative x^2 thats why i wasnt getting to required equation

the given solution has a positive

well

thanks

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[elder gazelle]

i dont get how thiss equates to 400e^-43 sqrt337

[elder gazelle]

another one of the same answer

[stone cipher]

do you know what the dot product is

[elder gazelle]

somewhat

[stone cipher]

ok

what's the dot product of those two vectors in the first question

[elder gazelle]

its -2 * -2/sqrt337 + 3*3/sqrt337 ... no?

[stone cipher]

yes

so what is that...

oh wait you're right that shouldn't be 400

[elder gazelle]

but thats the answer

[stone cipher]

wait no ok it works out

ok ok

what's 2^2 + 3^2 + 18^2

[elder gazelle]

337

[stone cipher]

ok

so it's 337/sqrt(337)

which is just sqrt(337)

so the dot product is just sqrt(337)

[elder gazelle]

oh

[stone cipher]

so 400 e^-43 * the dot product is just 400 e^-4e sqrt(337)

[elder gazelle]

why is it 337 its some rationlizing thing right

what is it called agfain

as in 337/sqrt337 = sqrt337

[stone cipher]

there's a thing called rationalisation but this isn't it

it's just division

i guess you could put it like

337 / (337)^0.5 = 337^0.5

power rules

[elder gazelle]

ah shit omg

thanks so much

.close

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[heavy knoll]

help whats next

the answer should be 35y+8 / 20xy

[static sable]

did you send us a picture of something you've rubbed out

[heavy knoll]

i wrote it using highlighter

.close

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[jagged sable]

i keep getting 3y-4

[swift rock]

show ur working

[jagged sable]

i did 9-16y^2 over 3-4y^3/y

i switched the ys cuz they can't be negative

then for the bottom fraction i multiplied 4y^2 by y for common denominator

[swift rock]

$$y^{-2} = 1/y^{2}$$

[clever fjord]

JustToPro

[jagged sable]

WAIT WHAT

[swift rock]

$$9-16y^{-2} = 9 - \frac{16}{y^{2}}$$

[clever fjord]

JustToPro

[jagged sable]

OHHHHH

i moved it to the other fractionnnn

wait let me try again

btw i have another qs

how is this not 5

[swift rock]

how are u getting 5?

[jagged sable]

5/3

yea

[swift rock]

in simplest form that is equal to N\M

simplify the fraction to the simplest form

and then the numerator is equal to N , not before

[jagged sable]

huh

wait

I STILL GOT 5

bro i knew i should've started studying earlier

i still have like 4 sections

[swift rock]

can u send a ss of ur working or anything?

im getting 25

[jagged sable]

45!?

okay wait

wait so 1+5/2x is N right

[swift rock]

no

"in simplest form" the fraction becomes equal to N / M

[jagged sable]

ohh

okay wait

[swift rock]

u gotta simplify the fraction before making them equal

[jagged sable]

OHHHHHHH

wait i'm so so so sorry

[jagged sable]

but i'm still getting the same answer for this

3y-4

i'll write down my work

wait

i think i got it

when -16 and -4 are cancelled out

[swift rock]

are the numbers in the boxes the actual answers?

[jagged sable]

does the 4 become positive

[jagged sable]

yea

[swift rock]

there is a identity being used there

$$a^2 - b^2 = (a+b)(a-b)$$

[clever fjord]

JustToPro

[swift rock]

u cant divide numbers when they are being added or subtracted

[jagged sable]

well what happens after

this part

factorize?

[swift rock]

$$9y^2 = (3y)^2$$
$$ 16 = 4^2$$

[clever fjord]

JustToPro

[swift rock]

use the identity i sent above

[jagged sable]

i'm so confused rn

[swift rock]

here u cant cancel out 9y^2 with 3y and 16 with 4 , not possible

[jagged sable]

do i factorize 9y^2 - 16

(3y-4) (3y+4)

OHHH

OMG I GOT IT

[swift rock]

$$\frac{9y^2 - 16}{3y-4} \neq \frac{9y^2}{3y} - \frac{16}{4} \neq 3y-4$$

[jagged sable]

THEN I CANCEL THE 2 (3Y-4)

[clever fjord]

JustToPro

[swift rock]

yeah like that

[jagged sable]

OMGMGMGMMGMGG

THANK YOU SO MUCH

THIS IS THE HAPPIEST MOMENT OF MY LIFE

I OWE YOU MY KIDNEY

.close

[odd edge]

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[odd edge]

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[mystic saffron]

How can I calculate the number of digits of 2010¹⁰?

[stone cipher]

the number of digits of a number n is log(n) but rounded up

log_10 (n), that is

[winged bear]

fun fact spotted

I will remember this

[dusty badger]

Is that a formula

[stone cipher]

sure?

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

Wait, what?

[stone cipher]

so take like 546

546 is between 100 and 1000

so log_10(546) is between 2 and 3

so if you round up it's 3

which is the number of digits

[mystic saffron]

Yes, but why does this work?

[stone cipher]

well it's sorta intuitive from the example

i mean let's make it rigorous

if a number has n digits

it is at least 10^(n-1) but less than 10^n

so log of it is between n-1 and n

so if you round it up, it's n

[coarse shuttle]

If you can prove that $\sum_{i = 0}^k a_i \cdot 10^{i-k}$ is inferior to 10 then you've got yourself the formula

for 0<a_i <10 of course

because : $n = \sum_{i = 0}^k a_i \cdot 10^i
\ \log_{10}(n) = \log_{10}(\sum_{i = 0}^k a_i \cdot 10^i) = \log(10^k(\sum_{i = 0}^k a_i \cdot 10^{i-k})) = \log(10^k)+\log(\sum_{i = 0}^k a_i \cdot 10^{i-k})=k+\log(\sum_{i = 0}^k a_i \cdot 10^{i-k})$

[clever fjord]

Azenx

[coarse shuttle]

where k+1 is the number of digits

[clever fjord]

Azenx

[mystic saffron]

I see...

Thank you so much

[coarse shuttle]

Btw to prove that the "reminder" is smaller than 10 and is positive you can just compute it for its max and min value

$\sum_{i = 0}^k 9 \cdot 10^{i-k}$

[clever fjord]

Azenx

[coarse shuttle]

And $\sum_{i = 0}^k 10^{i-k}$

[clever fjord]

Azenx

[coarse shuttle]

These two are power sums so you should be able to compute them

and you've proved the formula

[mystic saffron]

Some work to do for the rest of the day... 😆 Thanks

[odd edge]

@mystic saffron Has your question been resolved?

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[verbal dove]

x+1/x+2 < 1

[verbal dove]

Helpp!!
Anyoneee??

[frigid canopy]

!show

[odd edge]

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

[verbal dove]

x-1/x+2 < 1

Solve anyone

[heady raptor]

Can u calm down ?

we're humans dood

[verbal dove]

lol kk

[pastel orbit]

show what you’ve done so far

[heady raptor]

Why soo stubborn

[verbal dove]

I jsut asked for help
What is this server for anyways

If u aren;t gonna help me?

[pastel orbit]

!show

[odd edge]

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

[pastel orbit]

Where exactly are you stuck?

[surreal pulsar]

We are trying to help, but show us how far you've gotten on your own

Also do you mean (x-1)/(x+2) < 1?

Make sure you parenthesize the numerators and denominators to avoid confusion

[stone cipher]

this is not a free answers server

it's help

not 'do my work for me'

what have you tried?

[heady raptor]

• it's human who takes time to help u. You won't instantly get a response

[quasi sparrow]

!volunteers

[odd edge]

Helpers are just people volunteering their time to help you. Be polite and patient.

[verbal dove]

Yhh

[surreal pulsar]

OK, that's a little different. Your thoughts?

[verbal dove]

I crossed multipied

Which gives me x-1 = x+2

[surreal pulsar]

Remember, with an inequality, multiplication can change the direction of the sign

[verbal dove]

yh i know that

When dividing with - it changes

[surreal pulsar]

Correct. Also be sure to consider the special case where the denominator is 0 so you don't lose a possible switchover

[verbal dove]

Quadratic Equation?

[surreal pulsar]

No, (x-1)/(x+2) has a 0 denominator when x = -2, just consider that one of your critical points

[verbal dove]

0 as denominator??

[surreal pulsar]

Because it means the function approaches an asymptote and may change sign

Actually have you tried graphing (x-1)/(x+2)?

[verbal dove]

We haven't started graphical in skl

[surreal pulsar]

Right, but just on paper or something

[verbal dove]

Nope

I haven;t been thought tho

[surreal pulsar]

One sec

[verbal dove]

kk

[surreal pulsar]

https://www.geogebra.org/calculator/ycaxmhq6

Like that

[verbal dove]

k

[surreal pulsar]

Does that help?

[verbal dove]

Exactly what am I supposed to do?

[surreal pulsar]

Well, where is that graph < 1 and where it is bigger than 1 and why?

[verbal dove]

With that

[surreal pulsar]

Use that graph to help you see and then use the equation to get the exact answer

[verbal dove]

Can I say smth

[surreal pulsar]

Sure

[verbal dove]

We aren;t supposed to use graph

[surreal pulsar]

Oh, OK.

(x-1)/(x+2) < 1 -- subtract 1 from each side

That's way easier sorry

[verbal dove]

Done that alr

[surreal pulsar]

What did you get?

[verbal dove]

But when I put = 0 doesn;t that make it a quadratic equation?

[surreal pulsar]

Not at all

[verbal dove]

Bruh

[heady raptor]

The weird thing is that when u're doing the equation u find that the x delete themself

[surreal pulsar]

I have to be going, sorry, I'm hoping @heady raptor can help more

[heady raptor]

Leave it to me !

If x,y are real numbers and if $y = \frac{x - 2}{x + 1}$ , then $y(x + 1) = x - 2 \implies y(x + 1) + 2 = x$

[clever fjord]

Watynecc

[heady raptor]

So i got the same problem

Soo I wonder if we could actually facotrise

[verbal dove]

Yh

[heady raptor]

let me cook something

[verbal dove]

I thought of that

[verbal dove]

kk

[verbal dove]

u done?

[heady raptor]

yeah let me send the image

why tf my phone put it like this

[verbal dove]

The funny thing is that the answer is x > -2

I just need the workings

u there @heady raptor

[heady raptor]

yeah

sry mom called

[heady raptor]

There's 2 solution right ?

[heady raptor]

How u could if I got this while

maybe check the limit of (x-1)/(x+2) ?

[verbal dove]

kk

[verbal dove]

Not sure

[heady raptor]

Hm so x = 1

[verbal dove]

It's less than not less than or equal to

[heady raptor]

Oh my bad

[verbal dove]

Not sure we supposed to use graph

[verbal dove]

np

[heady raptor]

yeah... but at least we have some clues

[verbal dove]

yhh

[heady raptor]

Still don't work

[verbal dove]

Bruh

[heady raptor]

maybe we don't need to find x ?

[verbal dove]

Idk at this point just confused

[heady raptor]

(x-1)/(x+2) < 1 <=> 0 < 3 which is true ?

[heady raptor]

Yeah me too ;-; I hate fraction just for this shit

[verbal dove]

Exactly

[heady raptor]

I bet it's a missinput from the teacher ;d

[heady raptor]

Or else there's no x and we done with this

Alr I'm out i'm going to pray I hope I helped you ;-;

Take note for next class maybe it would be precious at the exam

[verbal dove]

It from a text book lol
Am gonna drag for bonus mark lol

[verbal dove]

lol

[heady raptor]

Oh.

[heady raptor]

hey ! ;-;

[verbal dove]

.close

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[steep thicket]

How would I be able to differentiate this?

[steep thicket]

.close

[odd edge]

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[potent folio]

you need to choose u such that its derivative is already there

so what can you pick that has a derivative you can already see?

what is the derivative of ln(x)?

yep! and look at that, it's already there :)

so choose u = ln(x)

ofc!! hopefully that helps

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@tiny brook Has your question been resolved?

[odd edge]

@tiny brook Has your question been resolved?

[odd edge]

@tiny brook Has your question been resolved?

[leaden quiver]

hello @tiny brook yes we need to evaluate ln|ln(x)| between x=e and x=6

so we get ln|ln(6)| - ln|ln(e)| = ln(ln(6))

yes it should be OK

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[static lance]

Is the solution correct?

[static lance]

I don't know a website to check for a curve

[quasi sparrow]

yea all 3 are right

[static lance]

thank you

.close

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[static lance]

for question d. g'(4) would be the original function at x = 4 which is -2?

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[near violet]

how do i do 1c and 1d?

[odd edge]

@near violet Has your question been resolved?

[sterile blaze]

from the first row you get, c2=-c1-2c4. repeat this for each row and try to replace where possible everything with just the pivot columns, c1, c3, c5, c7

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[lunar summit]

how do u do this

[odd edge]

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[lunar summit]

how do you do this

[sterile blaze]

new channel

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[open elbow]

is pearson's coeffecient same or similar to the r from a linear regression?

[odd edge]

@open elbow Has your question been resolved?

[open elbow]

<@&286206848099549185>

[brazen rune]

What

?

[valid pewter]

hm?

[open elbow]

is pearson's correlation coeffecient same or similar to the r from a linear regression?

[brazen rune]

idk

ask chatGPT

he knows everything

[valid pewter]

Pearson correlation measures the strength and direction between two numeric variables while simple linear regression describes the linear relationship between a response variable and an explanatory variable

[forest sky]

they are the same thing usually

like if you have a calculator that performs a linear regression and outputs a variable called 'r' or 'correlation' that's most likely pearson's coefficient

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[mystic saffron]

ok so

if 2 lines are perpendicular to each other their slopes must be reciprocals of each other

you can use this and point slope form to create the equation for the line, then draw it

if you are done with this type “.close”

which is?

[odd edge]

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[placid lynx]

what do you mean by this

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[tired obsidian]

I'm doing the Bolzano-Weierstrass theorem, but I didn't understand one thing: I know that A<an<B, by doing the proof I obtained 2 sequences Ak and Bk which tend to 2 limits. I didn't understand how the limit of the difference (B-A)/2^k which tends to 0 leads me to say that the two limits of Ak and Bk are equal.

[odd edge]

@tired obsidian Has your question been resolved?

[tired obsidian]

<@&286206848099549185>

[quasi sparrow]

show the entire proof you're referencing

[tired obsidian]

\textbf{Bolzano-Weierstrass Theorem}: Given a bounded sequence ${a_n}$, there exists at least one convergent subsequence.

\textbf{Proof}: Let's assume the sequence ${a_n}$ is bounded, meaning there exist constants $A$ and $B$ such that $A \leq a_n \leq B$ for all $n$ in the natural numbers.

We divide the interval $[A,B]$ by its midpoint $C = \frac{A + B}{2}$, creating two intervals: $[A,C]$ and $[C,B]$. At least one of these intervals contains infinitely many terms of the sequence ${a_n}$. In other words, either the set ${n \in \mathbb{N} : a_n \in [A,C]}$ or ${n \in \mathbb{N} : a_n \in [C,B]}$ is infinite.

Suppose, without loss of generality, that $[A,C]$ contains infinitely many terms, denoted by $[A_1, B_1]$, where $A \leq A_1$ and $B_1 \leq B$. This interval satisfies $B_1 - A_1 = \frac{B - A}{2}$.

We repeat this process with $[A_1, B_1]$, dividing it into two intervals $[A_2, B_2]$ and $[A_1, C_1]$, where $C_1 = \frac{A_1 + B_1}{2}$. Again, at least one of these intervals contains infinitely many terms of ${a_n}$.

Continuing this iterative process, we obtain sequences $A_k$ and $B_k$ such that:
\begin{itemize}
\item $A \leq A_k \leq A_{k+1} < B_{k+1} \leq B_k \leq B$, for $k$ in the natural numbers.
\item $B_k - A_k = \frac{B - A}{2^k}$, for $k$ in the natural numbers.
\end{itemize}

Moreover, the interval $[A_k, B_k]$ contains infinitely many terms of the sequence ${a_n}$ for all $k$ in the natural numbers.

This process guarantees the existence of a convergent subsequence of ${a_n}$.

[clever fjord]

alee

[tired obsidian]

In particular, the interval $[A_1,B_1]$ contains elements of the sequence $a_n$, thus there exists the first integer $n_1$ such that $a_{n_1} \in [A_1,B_1]$. Similarly, there exists a first integer $n_2$ greater than $n_1$ such that $a_{n_2} \in [A_2,B_2]$. By iterating this process, for $k = 1, 2$ and extending it to $k = 3, 4, 5, \ldots$, we establish a strictly increasing sequence of integers:
[ n_1 < n_2 < n_3 < \ldots < n_k < n_{k+1} < \ldots ]
such that $a_k \in [A_k,B_k]$ for every $k \in \mathbb{N}$.

Since $B_k - A_k = \frac{B-A}{2^k}$, we have:
[ A_k \leq a_{n_k} \leq B_k = A_k + \frac{B-A}{2^k} \quad \text{for } k \in \mathbb{N} ]

From equation (34.9), the sequence $A_k$ (and also $B_k$) is monotonic and bounded. By the theorem on monotonic sequences, $A_k$ admits a finite limit as $k \rightarrow +\infty$. Let's denote this limit as $\iota \in \mathbb{R}$.

Considering that $\frac{B-A}{2^k} \rightarrow 0$ as $k \rightarrow +\infty$, both the first and the last terms of equation (34.12) converge to $\iota$ as $k \rightarrow +\infty$. Therefore, by the Carabinieri theorem, we conclude:
[ \lim_{k \rightarrow +\infty} a_{n_k} = \iota ]

This expression represents the limit of the subsequence $a_{n_k}$ as $k$ tends to infinity.

[clever fjord]

alee

[tired obsidian]

@quasi sparrow

[quasi sparrow]

B and A are fixed, and the denominator grows as 2^k so the fraction goes to zero

[tired obsidian]

$A_k$ tends to $A_{\infty}$. $B_{k}$ tends to $B_{\infty}$. How are $A_{\infty}$ and $B_{\infty}$ related to each other?

[clever fjord]

alee

[tired obsidian]

idk that

<@&286206848099549185>

[quasi sparrow]

they don't relate and you don't need to know

[tired obsidian]

??

why

I claim that $A_{\infty} = B_{\infty}$.

[clever fjord]

alee

[tired obsidian]

$$ |B_{\infty} - A_{\infty}| \le |B_k - A_k|. $$

[clever fjord]

alee

[tired obsidian]

Apply squeeze theorem:

$$ |B_{\infty} - A_{\infty} | \le \lim_{n\to\infty} |B_k - A_k| = 0.$$

[clever fjord]

alee

[tired obsidian]

$$B_\infty \lim_{n\to\infty} B_n = \lim_{n\to\infty} (B_n-A_n) + \lim_{n\to\infty} A_n = \lim_{n\to\infty} A_n = A_\infty$$

[clever fjord]

alee

[tired obsidian]

That works, too.

@quasi sparrow is this reason correct?

<@&286206848099549185>

[quasi sparrow]

Why does this matter

[quasi sparrow]

you only need to know $\lim_{k \to \y} A_k = \iota$

[clever fjord]

riemann

[tired obsidian]

bc i use it

$$ |B_{\infty} - A_{\infty} | \le \lim_{n\to\infty} |B_k - A_k| = 0.$$

[clever fjord]

alee

[tired obsidian]

here

[quasi sparrow]

oh you're being confusing by a making new variable for one that already existed

yes they're equal

[tired obsidian]

okkkkkk

.close

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[shadow pilot]

i know i can use (a + b)^2 = a^2 + 2ab + b^2

but how would i distribute it the long way

i end up with 25 + 1v4

does the 5 make it 5v2?

[mystic saffron]

wait is the problem (sqrt(2) + 5)^2

[shadow pilot]

yes

[mystic saffron]

well so distributing it the long way would work like this

sqrt(2)*sqrt(2) + sqrt(2)*5 + 5sqrt(2) * 55

[shadow pilot]

wait im actualyl slow

i think ive done too much math today and need a break

i wasnt multiplying my 5's with my** 1**v2

[mystic saffron]

yeah

wdym v btw

[shadow pilot]

square root

[mystic saffron]

ok

[shadow pilot]

ty

.close

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[mystic saffron]

Hello I've been working on understanding compactness for a couple weeks now and I just can't seem to tie it all together. Ive been using baby rudin and many yt videos, but i cant connect the idea to continuous, compact maps on R are uniformly continuous. anyone have any tips for the intuition behind this?

[mystic saffron]

I knows theres a lot deeper concepts that play into what compactness is really meant to do, like discrete + compact -> finiteness, but i dont want to go into the rabbit hole of discrete math

[blissful tree]

what's the question, more specifically?

[mystic saffron]

how do you think of compactness

what is the intuition behind it

what does satisfying compactness mean to you

[blissful tree]

are we talking open cover definition?

[rigid dragon]

Every open covering has a finite sub covering

That’s what it means to me 😳

[mystic saffron]

either the open cover or sequential one

[blissful tree]

sequential one is less intuitive imo

[mystic saffron]

yeah but that doesnt tell you anything about anything

[rigid dragon]

It basically generalizes the notion of closed and bounded to general topological spaces

Any possible way that you cover the space with open sets you can choose finitely many of them and still cover the space

[mystic saffron]

I found the sequential one easier to imagine but the cover one is more useful

[rigid dragon]

The kicker is that it has to be for any possible covering

I think it’s kind of hard to intuit without just working through a bunch of proofs about it tbh

Which is a kinda disappointing answer I suppose

[mystic saffron]

i know what compactness is, Ive been studying it for two weeks now, I just want to connect it to everything else ik

should i learn more about connectedness

first

[rigid dragon]

Did you ever see the morphocular video on it?

[nocturne belfry]

honestly if you can convince yourself inverse image of an open set is open for a continuous function what you said plops out, i think

[mystic saffron]

yes

[rigid dragon]

I remember this helping me a bit

[nocturne belfry]

i went through this process 2 days ago

combine this fact with compactness definition with open covers

[mystic saffron]

Ive been looking at that recently, but I need to go over inverse images

[nocturne belfry]

its not so painful a proof

its really just element chasing

and using a couple balls

[rigid dragon]

Oof but that’s the definition of continuity

[mystic saffron]

yeahi dont know enough about images yet

[rigid dragon]

At least in the gigachad world of topology

[nocturne belfry]

you assume continuity

you want to show uniform continuity

right?

[mystic saffron]

yes

[nocturne belfry]

so you can use the definition

https://math.stackexchange.com/questions/843909/prove-that-the-inverse-image-of-an-open-set-is-open

first answer here made sense to me

maybe you too

[mystic saffron]

okay thanks

where did you learn about images and reverse images of sets

[nocturne belfry]

I am in a stupid graduate analysis class so i'm having to do a lot of review

[mystic saffron]

just scrolling through the wikipedia definition, it looks like something ive studied but with another name but idk

[nocturne belfry]

this was a good answer

see if you can disassemble what you want to show into a bunch of lemmas and a theorem

then work through them

just keep breaking stuff down until you reach something that seems elementary to you and then stop

[mystic saffron]

Is it really more efficient to just go into proofs without feeling fully confident in the idea?

[rigid dragon]

You probably won’t feel confident without proving it

[nocturne belfry]

IDK i can only speak for myself that sometimes there isnt intuition to be had prior to convincing yourself

[rigid dragon]

So I guess it could be a chicken and egg issue

[nocturne belfry]

sometimes things have a nice conceptual framework that you can work from

[rigid dragon]

But I would just dive into trying to prove lemmas for yourself

[nocturne belfry]

other times the framework is mostly that its aligned with a series of consequences

[nocturne belfry]

thats why i say this

[rigid dragon]

I’ve always found that my experience in analysis was a lot of definition chasing

[mystic saffron]

ill probably just try to prove things and it might come easier

[nocturne belfry]

ye

[rigid dragon]

Just directly restating all the definitions makes it sorta easy

[nocturne belfry]

you just need to be honest with yourself

start reading the proof of the thing you want to understand

if they say something you dont believe then you have work to do

[rigid dragon]

At least at first, measure theory was a different beast. Just have to come up with some whack sets out of thin air for that one

[nocturne belfry]

i say all this bc i went through heine borel as a person who barely passed real analysis on thursday

it took like 12 square feet of whiteboard space

[rigid dragon]

Heine Borel theorem is kinda the motivation behind compactness in some sense

It allows you to expand the notion of closed and bounded sets to topological spaces in general

[mystic saffron]

self studying this all is slow

[nocturne belfry]

its a benefit, not a curse

you have time to build as strong of a foundation as youd like

[mystic saffron]

at least ill have a strong basis

yeah

ill work through the image things thanks guys

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[lament stone]

What is the area between the sine and cosine waves between the interval pi/4 and 5*pi/4? (The purple area is what I need to find the area of.

[lament stone]

I just don't know where to begin.

[mystic saffron]

Just to make sure: The signed area or the geometric area?

[lament stone]

The geometric area.

[mystic saffron]

also what sine and cosine waves are they. Can you share both functions

[lament stone]

They're just normal sine and cosine waves. Nothing special about them.