help-4

[marble arch]

With 2

[wintry cape]

👍 so how does all of this help me move forward from this

[marble arch]

There'd be an (x - 5) multiplied to the 2x

[wintry cape]

Alright

So make another bracket with that?

[marble arch]

Yes, (2x(x - 5) - 4x²)

[wintry cape]

Just to be sure I keep the other stuff too or no

[wintry cape]

Just his

[marble arch]

Yes yes, keep it outside

[wintry cape]

This

[marble arch]

The (x - 5)³

[marble arch]

Yes that's correct

Now simplify 2x(x - 5)

And evaluate it with -4x²

[marble arch]

Yeah that's fine, now you can take -2x common from (-2x² - 10x)

So you can take it out

[marble arch]

Whoops, it'd be +5, we've already taken the minus sign outside

[wintry cape]

Silly mistake 😭

Ok is that the final answer
I think it is

[marble arch]

Yes it is

[wintry cape]

Anything I can write to show how I went from the first part to the second

I know that I will look back and be confused on how I went from point a to point b

[marble arch]

Well, the first step shouldn't be there at all

The second step should be the first step

Also if it makes it easier for you, then you can make the (x - 5) inside as (x - 5)¹

[wintry cape]

(2x(x-5) - 4x^2(x-5)^3) is the first step

Then

Right

[marble arch]

Yes, also close the bracket after 4x²

[wintry cape]

Wdym

Isn’t 4x^2 with the (x-5)^3

As that’s how it was in the problem

[marble arch]

Well, you've taken (x - 5)³ common

Not 4x² too, 4x² remains inside, like that's why we take something common in the first place

[wintry cape]

So like that then

[marble arch]

No mate, the bracket is after 4x² and before (x - 5)³

Sorry mate, I gotta go now

It's getting pretty late here

[wintry cape]

That’s alright thank you I was just tryna make sense of what u meant here

[marble arch]

Alright mate

[wintry cape]

Im js mathed out thanks for helping out being so patient

[marble arch]

No problem dude, maybe consider taking a break

[wintry cape]

Kinda hard when I get assigned so math work a day half of it my teacher explains horribly

[marble arch]

Damn man, well we're here to help you, and there are people better than me here, but still consider taking like a 15 min break or something

[wintry cape]

Yeah still got 5ish problems left and I will Ty

U had a lot of patience today that I don’t have

😭

[marble arch]

Well I shouldn't be helping people if I'm not patient :)

[wintry cape]

For that I thank you
Have a good rest of your night

[marble arch]

Have a nice one yourself mate

[wintry cape]

Fixed my step 1 with the brackets n stuff

Time for number 19
I should start first by looking for a GCF

Where I’m at

First bracket is a sum of two cubes
Second bracket is a difference of two squares

?

Can someone let me know if I did this right?

Number 19

[vale dock]

@wintry cape Has your question been resolved?

[wintry cape]

<@&286206848099549185> ?

[normal hollow]

Yes it's alright

But did u need to expand the sum of cube and sum of squares?

[wintry cape]

Omg I can go further still 😭

Fed up with factoring so don’t mind me

[normal hollow]

But do u hv the choice to not expand this and keep it as is for an answer
2y(27x³+y³)(9x²-4y²)

[wintry cape]

I circled my answer

[wintry cape]

Look under that

I did go further

[normal hollow]

I know that u went further and u did right

[wintry cape]

Really 😭 thank you
I’m sorry I am really exhausted with this worksheet

[normal hollow]

U dont hv any reason to be sorry

Get some rest

[wintry cape]

Right after I finish these last 2

Let’s attempt to do them on my own first

[normal hollow]

Good luck on that

[wintry cape]

Turns out I’m forgetting how to count

3 problems left

Did I do number 21 correct

[normal hollow]

U r forgetting the y⁵ at the beginning

[wintry cape]

That’s all I forgot right

Yippee now I can say dos problems left

I probably did this wrong but I’m unsure

Where to go with this

Number 22

[normal hollow]

Notice that u can take 2(x+4)⁵ common from each term

[wintry cape]

Ok what do you do with the left ^2

[normal hollow]

Expand,most likely

[wintry cape]

Explain more on this 😭

[normal hollow]

(x+y)² formula

(x+4)²=x²+2.4.x+16

this is wrong btw

[wintry cape]

Had a feeling

How would I go about it then

[normal hollow]

$2(x+4)^{7} -36x(x+4)^{5} =2(x+4)^{5}[(x+4)^{2} -18x]$

[rocky lotus]

vικτιμιζερ

[normal hollow]

$2(x+4)^{5}[(x+4)^{2} -18x]=2(x+4)^{5}[x^{2}+2.4.x+16-18x]$

[rocky lotus]

vικτιμιζερ

[normal hollow]

Now factorize the 3rd bracket portion

[wintry cape]

Ok so start with this

[wintry cape]

What happened to the -36

[normal hollow]

2×18=36

I took 2 common as well

[wintry cape]

So then wouldn’t the 2 become 1

I’m confused

[normal hollow]

Can u mark the part u dont understand

[wintry cape]

You made the 36 into 18
Wouldn’t the 2 also change to just 1

[normal hollow]

It change to 1

Lemme write it down ig

[wintry cape]

👍

[normal hollow]

See if u get it now

If not mark it

If yes,then proceed with factorizing x²-10x+16

[wintry cape]

Is that a 2

[normal hollow]

Yes

[wintry cape]

I still don’t really see how you went from this to that
And also what do the 3 lines supposed to show

[normal hollow]

Squared

2 and (x+4)⁵ r the GCFs

[normal hollow]

Do u see that

[wintry cape]

And (x+4)^2 is?

U have a line under that too

Tbh I’m a lot more confused now 😭

[normal hollow]

Okay see 5xy-10x²y
The gcf is 5xy
So u write 5xy(1-2x)

That's what i m doing

Do u know this that x⁵.x⁶=x^(5+6)=x^11

It's the same (x+4)^7=(x+4)^(5+2)=(x+4)^5 .(x+4)^2

I took the (x+4)^5 common
Which leaves a (x+4)^2

[wintry cape]

. You mean multiplying right

[normal hollow]

Yes

. Or ×

Both r multiplying

[wintry cape]

Ok

I see

U just broken it up

But going back to the 36 and 2

[normal hollow]

See 2y -36x
Gcf is 2
So 2(y-18x)

Right?

Similarly 2(x+4)⁷ -36x(x+4)⁵
=2(x+4)⁵[(x+4)²-18x]

[normal hollow]

There is no 2 in front of (x+4)²

[wintry cape]

Got it okay

[wintry cape]

I know your saying you just did 2(y-18x) and it makes sense like that but nothing is written like that in what you broke down here

[normal hollow]

Okay note that the y in the example i showed is (x+4)² ,now do u see anything?

I think u should try to solve it urself,and then u will understand

[wintry cape]

Which example

[wintry cape]

I’m so lost on solving it myself 😭

[normal hollow]

The 2(y-18x)

[normal hollow]

🥲

Take rest,do this later

[vale dock]

@wintry cape Has your question been resolved?

[wintry cape]

I need to finish this 😭

But I really wanna understand what I’m doing

[rigid pivot]

What do you need help on?

[wintry cape]

Number 22

[marble arch]

take 2(x + 4)^5 common

[wintry cape]

I’m looking back is this basically what im doing

[rocky lotus]

xfilin

[wintry cape]

These a class notes btw

[rigid pivot]

This is what you need to do

I’ll give you a better explanation

[rigid pivot]

To simplify the expression $$2(x+4)^{7}-36x(x+4)^{5}$$, we first look for common factors.Identify the common factors: Both terms have a factor of 2 and a factor of $$(x+4)^{5}$$Factor out the common terms: We pull out $$2(x+4)^{5}$$ from both parts of the expression.$$2(x+4)^{7}$$ divided by $$2(x+4)^{5}$$ leaves us with $$(x+4)^{2}$$$$-36x(x+4)^{5}$$ divided by $$2(x+4)^{5}$$ leaves us with $$-18x$$So, the expression becomes $$2(x+4)^{5}[(x+4)^{2} - 18x]$$Simplify inside the brackets: We expand $$(x+4)^{2}$$ which is $$x^{2} + 8x + 16$$Now, the expression inside the brackets is $$x^{2} + 8x + 16 - 18x$$Combine like terms: We combine the 'x' terms: $$8x - 18x = -10x$$The expression inside the brackets is now $$x^{2} - 10x + 16$$Factor the quadratic: We need to find two numbers that multiply to 16 and add up to -10. These numbers are -2 and -8.So, $$x^{2} - 10x + 16$$ factors into $$(x-2)(x-8)$$Final Answer: Putting it all together, the simplified expression is $$2(x+4)^{5}(x-2)(x-8)$$

[wintry cape]

[rocky lotus]

xfilin

[rigid pivot]

Brodie💀 I think I’m locked in

[wintry cape]

How do you type so fast..

[rigid pivot]

Typing club 😭

[rigid pivot]

3 minutes is a long time,
But anyway What will the answer be then?

[wintry cape]

How do you go from there to there

[wintry cape]

Welcome back

[marble arch]

Dude did you take a break yet😭

[wintry cape]

I have raged a lot

Does that count

[marble arch]

That's the opposite😭

[wintry cape]

And I’ve went do a lot of other work

Cus I’ve been on the same math worksheet for hours

And my teacher is practically impossible to understand and deal with

[wintry cape]

Oh well

[wintry cape]

@rigid pivot @marble arch

[rigid pivot]

Yes?

[marble arch]

I've honestly got no clue dude, who wrote it?

[wintry cape]

Also can you guys see this class notes and see if that’s what we are doing here
Maybe I learned it another way and you guys are showing it another and I’m confused that way

OR

it’s just a complete different topic and I’m sleep deprived

[wintry cape]

Look at what I said there

I was tryna process your break down

[wintry cape]

But how did you go from 2(x+4)^5 to ^7 part

[rocky lotus]

xfilin

[rigid pivot]

Do you understand it now or do you not get something?

[wintry cape]

I’m still on abt the -18x bc Im not grasphing that- how did you get that

[rocky lotus]

xfilin

[rigid pivot]

What do you think?

[wintry cape]

👍

We are left w ??

[wintry cape]

Did we divide the whole equation by the common term?

[normal hollow]

Wow r we on the same math now

[wintry cape]

Huh

[normal hollow]

R u done with 22

[wintry cape]

Still are on the same equation.

If that’s what u mean

I think I’ve figured out this is the same thing as u substitution

You guys were showing it to me a different way then I learned it

[normal hollow]

Oh so plug in x+4=y

[normal hollow]

Doesnt hurt to learn both ways

[wintry cape]

Yeah I did all of that

[wintry cape]

Yeah of course

But that’s why I was so unfamiliar

[normal hollow]

Good so far,factorize the last portion as well

[wintry cape]

Yep working on it

[normal hollow]

I get it
That thought crossed my mind but i thought would be increasing steps for no reason

[wintry cape]

Can’t thank all of you guys enough omg

For your helpfulness and support

I appreciate it plenty

[normal hollow]

At the end it was u who solved it tho

[wintry cape]

😭 still

I hope to develop the amount of patience you all have

[marble arch]

Yes dude that looks correct

[wintry cape]

I did the last one

Am I done

Pls tell me i done

[marble arch]

Yes, you're done, that's correct!

NOW GO REST!!!

[runic scroll]

just coming in to say, well done!

[wintry cape]

All of you are amazing 😭 thank you so much

I am happy to say

U can close this ticket

[marble arch]

You gotta do it mate

[wintry cape]

Oops 😬

[marble arch]

It's fine

[wintry cape]

Till I see u again 👋 @marble arch @normal hollow

.close

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[river arrow]

a most curious math problem appeared on the server some time ago and filled me with zeitgeist. i finally found an equation for the answer but im not sure how to unravel the trigonometry. any hints would be appreciated orz $\frac{\sin(3x)}{\sin(135-3x)} = \frac{\sin(x)}{\sin(45-x)}$

[rocky lotus]

Plvzfq_rit

[river arrow]

(have to solve for x)

[marble arch]

Have you tried opening sin(3x)

[river arrow]

like thru a trig identity?

[marble arch]

Yes, sin3x = 3sinx - 4sin^(3)x

unless I'm mistaken

[river arrow]

i havent yet no

[marble arch]

Whoa, that might just do it I realised

[river arrow]

hmm lemme try that, as well as open out the other stuff

[marble arch]

Alright sure, let me know once you get it (which you should unless there's a gimmick or I'm missing something)

[river arrow]

ok... i think i mightve gone astray and did something wrong with my calculations

first i tried to expand the stuff at the bottom

$\frac{\sin(3x)}{\sin(135)\cos(3x)-\cos(135)\sin(3x)} = \frac{\sin(x)}{\sin(45)\cos(x)-\cos(45)\sin(x)}$

[rocky lotus]

Plvzfq_rit

[river arrow]

then those constants have special values, so i could do...

$\frac{\sin(3x)}{\cos(3x)+\sin(3x)} = \frac{\sin(x)}{\cos(x)-\sin(x)}$

[rocky lotus]

Plvzfq_rit

[river arrow]

hmm from there i tried expanding the left hand side with the reduction formula so...

[marble arch]

No no

sin(135 - 3x) = sin(3(45 - x))

That's also sin3, well, let's say y, that's also sin3y

[river arrow]

oh! yes

dang i didnt realize that, thanks

lemme try this again

[marble arch]

No problem mate

[river arrow]

hmm

should i have gotten $\sin^2(x)=\sin^2(45-x)$ in the end?

[rocky lotus]

Plvzfq_rit

[river arrow]

hmm i think i got it

yep this should be good

thanks again doctorstrangejr!

.close

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[scenic zealot]

Can I show a causal link between data using a line graph?

[turbid summit]

you mean linear regression?

[scenic zealot]

nah just have to prove that there is a link between hours of studying and the grades using "appropriate visualisations"

[turbid summit]

you use regression for that...

but corelation is not causation so i'm not sure

[scenic zealot]

ya but I need to show using an appropriate graph "Create appropriate visualisations (e.g., scatter plots, histograms, etc)"

[vale dock]

@scenic zealot Has your question been resolved?

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[lime dome]

it was supposed to be a 3 letter code and i thought the one for bifid was the easiest. However, i think i messed up since i got tabonexo myosorfs for the qclxqjpo poojjsrc

[lime dome]

im stuck on deciphering this

[sour needle]

First part just looks like binary numbers that presumably correspond to the alphabet

[lime dome]

Ohh what i did was converting them to denary then counting the individual letters 😭 maybe thats why that part went wrong too

[sour needle]

110 = 6,
1 2 3 4 5 6
a b c d e f

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[lusty bay]

ignore hte numbers and letters (its from a random youtube video)

the circle on the left is split into many (proper) triangles like shown on the left. then its rearranged to from that rectangle... yeah, if hte radius is 10, how much longer will the rectangle be from the circle?

so he wants to compare between the diameter of the circle and the longer length of the rectangle. the diamter of the circle is 2r., and the longer length of the rectangle is basically the circumference of the circle (this is my thought process, not stating facts)

so that should be 2 pi r - 2r ---> 20 pi - 20 --> which has pi...? the answer doesnt have pi in it

[lusty bay]

or thats how it was shown in the video.

the answer is like... 20?

where did i go wrong

[stark wedge]

the answer doesn't have pi in it

uhhh what. 20???

you're gonna have to link the video cause you're not making any sense rn

[lusty bay]

I changed the given information so ignore the 20.

[stark wedge]

the rectangle, if we imagine its imperfections to be small enough not to matter, has height r and length pi r

[fickle rose]

it sounds like the definition of pi to me?

[jovial edge]

Maybe they round up the answer so pi isn't in here?

[lusty bay]

the video is random and embarassing nobody judge me 😡.

[stark wedge]

i mean we're trying to get to the bottom of your confusion not judge you

[pine prairie]

why do you think the answer has no pi in it?

[lusty bay]

the video

give me one second

ill link it

im just finding the proper timestamps

15:22 for the question
18:01 for the right answer (10)
i changed the context from diameter to radius but that shouldnt add a pi

https://youtu.be/VMHCyzTP4oE?si=szwcVd-3PGYnCl6h

if you comment on the choice of video

i will find you,

[lusty horizon]

Okay

[stark wedge]

hm

ok so the English of this problem ain't so great

but what i THINK is happening

is that they're asking for the difference between the shapes' perimeters.

[lusty bay]

would that justify the existance of pi

[stark wedge]

and this difference, to put it simply, consists entirely of the 2 vertical sides of the rectangle.

so the answer really is 2r

[pine prairie]

um

did you watch the explanation

[lusty bay]

i didnt understand.

nor did i comprehend how he started it.

[lusty bay]

the perimeter of a rectangle...?

[stark wedge]

i... dont understand what is unclear

[pine prairie]

you got xy'd in a video

lmao

[lusty bay]

cant we consider the perimeter of the rectangle to be the hypotenuse of two lengths r and 2pi r

that still gives pi

[pine prairie]

hypotenuse of the 2 lengths r and 2 pi r?

[lusty bay]

like a right triangle in the rectangle

[stark wedge]

what hypotenuse

[pine prairie]

there are no right angled triangles in this problem

[lusty bay]

from a right triangle that splits the rectangle in half

[pine prairie]

afaik there arent even triangles in this problem

[lusty bay]

the angles of a rectangle arent 90s?

[pine prairie]

that's not the problem

these 3 sided shapes dont have 3 straight sides

[pine prairie]

look at what the kid draws

[lusty bay]

I'm sorry but what exactly are you talking about?

[pine prairie]

see how his sides are bumpy

[lusty bay]

OH

oh...

yeah we dont straighten up the circumference of the circle

[pine prairie]

he claims that the 2 bumpy sides are just the exterior of the circle

[lusty bay]

nice.

[pine prairie]

so they will be just the same as the circumference in length

so the difference will just be the extra 2r pieces on the 2 ends

so the differences is 2r = d = 10

[lusty bay]

oh yes i messed up on that its only pi r for each

[pine prairie]

yeah that's what the kid said

im just repeating what he said lol

[lusty bay]

the kid spoke too quickly.

[pine prairie]

it helps that i understand whats hes saying rather than the shitty subtitles

[lusty bay]

why did they give him this misleading shape

[pine prairie]

ya the question was bad

[lusty bay]

i understand what comes after this

but not it exactly

wdym the diameter is equal to the circumference in length

isnt that the whole point of pi that they arent

[pine prairie]

i did not say that

[lusty bay]

my bad! please elaborate

[pine prairie]

i said

the 2 bumpy sides of the rectangular looking shape is just the exterior of the circle

so they have the same length

hence the difference will be the 2 ends of the rectangle looking shape

which has length r on each side, so together it's 2r = diametre = 10

[lusty bay]

oh my god i thought perimeter meant diameter

but for rectangle

this was all a miscommunication bc english si trash

[pine prairie]

yes

[lusty bay]

yeah now it makes sooo much sense

ty

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[jovial edge]

You can beat the kid keep going

[river shale]

[short solar]

isnt this frame from a video , i remember watching this a while ago , like a compedition between a 5 y old chinese? vs older?? 😭

i just scrolled up , im sorry for the ping

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[light egret]

Question:
If I have the set A = {1, 2, 3} and I have the relation R = {(1, 1), (2, 2), (3, 3), (1, 3)} (hope my notation is correct), does that mean the relation is reflexive? Or is it not reflexive because we have that (1, 3) in there?

[light egret]

It was something that confused me during lecture

[keen tundra]

reflexive means it contains (a, a) for every a

it doesn't make sense to disallow any other pair to be in the relation. As this forces any reflexive relation to be the identity

[light egret]

Okay that was what I thought too, thank you

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[rocky lotus]

brandon

[drifting hornet]

it's an if and only if, so it depends on which direction you do first

if you wanna start with the -> direction, then you assume that phi is an automorphism and prove that G must be abelian and has odd order

you probably only used the fact that phi is a homomorphism for now, so now its porlly time to use the fact that its also an automorphism

what does that mean? What can you infer about phi?

into itself

but thats obvious in this case

the important part is bijective

try using the fact that phi is bijective, to prove that G must have odd order

might be useful to look at some groups of even order and find out why squaring isnt bijective for them
(ping me if u need me)

indeed

or wait

if such element exist, then it's not bijective

but i'd go for contradiction here. Assume it's even, prove that such g exists, get a contradiction (phi is not injective) and conclude it's odd

lagranges theorem is more relevant for the other direction tbh

at least i think so

do you know what order of element is? Do you know that it divides the order of the group?

[keen tundra]

minimal n

[drifting hornet]

and do you know that order of any element g divides the order of the group G

or that order of any subgroup of G divides order of G itself

oh now thats even better

think about how u could apply it here

and think about how this relates to order of element (how u could rephrase it in terms of orders)

soo...?

yeah, and as you said before, g^2 = e gives you contradiction (bc that means phi(g) = g^2 = e = e^2 = phi(e), which contradicts injectivity)

and so the first dir is done

yep

the fact that its a homomorphism is quite trivial, so the only difficult part is gonna be the bijectivity

good thing is that for functions G -> G for finite G, injectivity = bijectivity = surjectivity, so for the other part you're just gonna have to prove injectivity

well, to start, id try reducing it to something^2 = e, youve already dealt with that once and its generally a bit nicer

[light saddle]

(i.e. always look at the Ker to prove a homomorphism is injective)

[drifting hornet]

do you know what a kernel is?

kernel is basically the set of all inputs to a function, which result in e when plugged in

i.e. {x | f(x) = e}

and to prove that a homomorphism is injective, it suffices to prove that the kernel is trivial (i.e only contains e itself)

but if this is all new to you, dw about it

[drifting hornet]

just do this

and once you're done, dig deep in your memory or notes and try to recall the lagrange's theorem. Im fairly sure you already know it, since you know cauchy's theorem.

Lagrange's theorem says either one of the following:

If H is a subgroup of G, then |H| divides |G|

or

If g is in G, then ord(g) divides |G|

[keen tundra]

to elaborate on why is that: f(g)=f(h) if and only if f(g)f(h)^-1 = e and since f is an homomorphism this is if and only if f(gh^-1)=e <---> gh^-1 in ker(f)

[drifting hornet]

this is usually proved very soon after introducing subgroups

yeah, thats essentially it

since G has odd order x=e
This is the step which uses lagrange's theorem btw. G cant have element of order 2, because then 2 would divide |G|, but 2 cant divide odd number

>>> something

or just > something for one liners

np

yep

you're welcome :)

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[marsh lagoon]

Please do not provide complete answers to helpees; and do not use AI generated work as your own to answer other's questions

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[distant galleon]

claim

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[amber gazelle]

ummm another google sheet formula question welp sorry lol

[amber gazelle]

Option 1 works, but I can't make option 2 to work

for example

works

but if I try sum countif it only reads -1 not 1 cells

[hardy coral]

Where is this from exactly?

[steady charm]

the formula works in excel, but google sheets does not have exactly the same syntax as excel so it doesn't work

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[fast bridge]

sqrt(x)(2x^2+3(x-(2x+1)))

I'm not sure how I can simplify the above problem, mostly because the parentheses seems to soft lock me into the things I'm allowed to do to equation.
I've managed to get it down to

sqrt(x)(2x^2+3x - (6x+1)

by distributing the 3 but I dont know where to go from here

[hazy pivot]

!original please

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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.

[hazy pivot]

Just to be sure that you're writing it correctly for us

[hazy pivot]

Cool

Let's start from the innermost expression

What's $x - (2x+1)$

[rocky lotus]

Xavier 🌺

[fast bridge]

mmm thers nothing i can do with this here right, I mean maybe distribute a negative 1?

[hazy pivot]

Yes

So what does it give you

[fast bridge]

x -2x -1

[hazy pivot]

Good

Now can you simplify that further

[fast bridge]

-x-1

[hazy pivot]

Good

Now what's $3(-x-1)$

[rocky lotus]

Xavier 🌺

[fast bridge]

-3x-3

[hazy pivot]

Good

What's your expression now

[fast bridge]

sqrt(x)(2x^2-3x-3)

[hazy pivot]

Good

Now there's only possible thing to do

[fast bridge]

I could distribute the sqrt of x inside

I don't think that makes sense actually

[hazy pivot]

Yes

[hazy pivot]

I mean you could leave it like it is too

What's your end goal here

[fast bridge]

simplify the expression

[hazy pivot]

I hate that sentence cuz it's not explicit

I'd call what you have here simplified enough

But sure you could distribute the √x

[fast bridge]

Alright, thanks, it really wasn't as confusing as I thought it would be

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[regal topaz]

I feel like I did something wrong

[regal topaz]

,rotate

[regal topaz]

heres the question

can someone explain what i did wrong or how i can get to the right answer

[modest sinew]

U pulled out dy/dx from (y^2 + 2xy*dy/dx + 4*dy/dx) even though the y^2 part doesnt have dy/dx

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[green steeple]

Hey guys I need some help understanding this

[green steeple]

Its asking to evaluate each limit with power rule

[flint phoenix]

So what's the context of power rule, do you know?

[green steeple]

You bring down the exponent to the coefficent and if there is one already you multiply them together and subtract one from the exponent

[leaden rivet]

Are you evaluating the limit or tryna find the derivative?

[runic scroll]

sorry, accidentally hit my keyboard.

[green steeple]

Says to evaluate

[leaden rivet]

Just evaluate? Nothing specific?

[fossil mulch]

From the first principle definition of a derivative this limit is equivalent to f'(5) where f(x)=x^(1/3)

That's question 32 no?

[green steeple]

yes

[fossil mulch]

Ain't this qn 31

[green steeple]

oh snap my bad i got confused

sorry for sending the wrong one

[fossil mulch]

So yea clue here

[green steeple]

f prime of x is 1/x^3 right?

[fossil mulch]

No

$f(x)=x^{\frac{1}{3}}$

[rocky lotus]

denzio321

[fossil mulch]

Power rule:$f(x)=x^n$ $f'(x)=nx^{n-1}$ where n is a constant

[rocky lotus]

denzio321

[green steeple]

so f'(x)= 1/3x^-2/3?

[fossil mulch]

Yes

Then just sub x=5 in there and do some simplification

[green steeple]

What went wrong

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[north scarab]

This is incorrect apparently

[north scarab]

answer key:

[stark wedge]

you forgor the chain rule

or rather you didn't differentiate x^(1/2) correctly for it

[north scarab]

let me try again

[stark wedge]

derivative of x^(1/2) is not "just 1/2"

[north scarab]

1/2-1

[fossil mulch]

Yep

[north scarab]

thank u I

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[mint slate]

apparently the answer is 17, how to get there

[mint slate]

am i tweaking or drawing that equal 18?

or am i geeking

oh nvm

i am

tysm sir

and or mam

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[final swift]

Did I set it up correctly?

[pure pulsar]

yep

[wild linden]

That looks correct to me

[pure pulsar]

Where’s the final answer

[pure pulsar]

I’d suggest doing cos^2 x - sin^2 x identity

[final swift]

oh like 25(cos^2(x)-sin^2(x)) = 25(cos(2x))

i get -20pi

[pure pulsar]

Yeah

Do that and then solve

[pure pulsar]

Solve the whole thing and lmk if you get stuck again

[final swift]

I got -20 pi

oh im stupid i wrote pi not pi/4

sorry

[wild linden]

Not stupid. Just learning.

[pure pulsar]

Chill, not a big deal, happens with all of us

[pure pulsar]

Can i correct one more thing?

[final swift]

yeah

[pure pulsar]

I think its 10cosx

[final swift]

ah

i got the right answer yay

it took me 4 tries to do this question man

thanks

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[pure pulsar]

All good

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[fathom narwhal]

how to solve the konesinberg bridge therrrom

[fathom narwhal]

Theorem

[coral urchin]

eulers criterion

that there exists an eulerian path <==> exactly 0 or 2 vertices have an odd degree

[thin valley]

I think he means how to solve the particular case the Königsberg problem. The euler's criterion is the general theorem to prove to solve it so applying it without reasoning is an act of faith in this case. I suppose he is interested in the proof of the criterion

[coral urchin]

i believe it comes from handshaking lemma and def of eulerian path

[thin valley]

So first thing you have to realise is that you can represent the problem with a graph

The problem is solved if you can find a path on the graph that passes through every edge just once finishing in a distinct or the same vertex

Are you there?

So suppose a general graph and a path on that graph for which that holds. The vertices in the middle can only be connected with an even number of edges (even degree) because when you go into one of them you always go out. This means the only vertices that can have an odd number of edges are the first one and the last one (two vertices of the graph)

A graph is eulerian if and only if it has 0 or 2 vertices with odd degree. (If it has two, every possible path goes from one of such vertices to the other).

Every vertex of the Königsberg has an odd degree, so one cannot find an eulerian path on it

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[final swift]

The answer should be 81pi/5 for the volume

did i set it up wrong

[cerulean scaffold]

Could you post the question as is?

[coral urchin]

81pi/5 is correct

although, if you use shell method, you dont have to covert everything wrt y

OH youre saying you didnt get 81pi/5

sorry

so, the shell method says that if we rotate about $x=h$, then $$V=2\pi\int_a^b(h-x)f(x)dx$$ for $a<b\le h$

i would use this method instead of washers and chasing y around

[rocky lotus]

Cycadellic

[final swift]

i think we didn't learn this one

[coral urchin]

hmm

but this is the entire setup to a shell integral

i mean, we can do washers anyways

[final swift]

if i use the shell method is it $$V=2\pi\int_0^1(2-x)(27x^3)dx$$

[rocky lotus]

Ana

[coral urchin]

yes

[coral urchin]

also, if you want to stick with washers, your setup is wrong

we subtract 2, not add it

[final swift]

why subtract

[coral urchin]

so the line we rotate about is moved to 0

if that makes sense

then its just usual washer from there at 0

hence $$V=\pi\int\left(((h-R_0)^2-(h-R_1)^2\right)dx$$

[rocky lotus]

Cycadellic

[final swift]

i see

[fathom narwhal]

.reopen

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@final swift Has your question been resolved?

[final swift]

i tried both methods but i get a wrong answer for the washer method even tho i should get same answers

[coral urchin]

Why is your upper bound 8 now?

27 was correct

[final swift]

this is another question, same type but different numbers

[coral urchin]

Oh

The setup is correct

The problen is the second step

You say
3-y^(1/3)+1/4 y^(2/3)
when it should be
3-2y^(1/3)+1/4 y^(2/3)

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[turbid pike]

can someone explain why this is logically true?

[stark wedge]

wym "logically"

this has very little logic or reasoning, it's just bashing

[turbid pike]

the pr0of is just expanding it

[stark wedge]

sure is.

[turbid pike]

thats it?

i thought it had some reasoning too

[stark wedge]

that's it.

[coral urchin]

I mean sure, algebra itself is a logical argument

[jovial edge]

maybe you want a combinatorial proof ?

[turbid pike]

sure ig

[jovial edge]

I meant do you really want that since combinatorial proof can get a bit tricky compare to the algebra one

[turbid pike]

then no need

still thanks tho

[turbid pike]

can i find it in youtube>

[rocky kettle]

lol i just proved this problem 1 week ago in math lesson xD

[jovial edge]

I'm not sure

[rocky kettle]

you should've seen my classmates face

[rocky kettle]

its actually easier

you just need pascal triangle for that

[turbid pike]

can you please

[rocky kettle]

logically is just pascal triangle

[limber cliff]

Why u need any reasoning?

[rocky kettle]

have you learnt pascal triangle?

[turbid pike]

just curious

[jovial edge]

Yeah...

Given a set A={a,b,c,.....}

[turbid pike]

no

[jovial edge]

this set has n+1 elements

I choose r elements in this set

[rocky kettle]

ok then

[jovial edge]

I have rCn+1

[rocky kettle]

its a triangle of which all the lines depends on the previous one

essentially just the 2 numbers above added together

[jovial edge]

On the other hand, I divide this into 2 cases

[turbid pike]

who shd i listen to?

[jovial edge]

1 that in r elements i choose, there is element a

[rocky kettle]

i think mine is simpler

[jovial edge]

I'm talking about a proof atm

[rocky kettle]

ok you first then

[jovial edge]

Yeah but it requires pascal triangle

[rocky kettle]

ping me when you're done

[turbid pike]

tbh not interested in knowing the reasoning now

anyways thanks and osrry you two

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[jovial edge]

that fair

[turbid pike]

just dont want to waste your time

.reopen

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[turbid pike]

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[midnight pier]

Where did I messed up

Answer was suppose to be 1/2

[keen tundra]

(1-sqrt(u))(1+sqrt(u)) is not 1-u^2 its 1-u

[midnight pier]

Omg no way I just did that mistake

Embarrassinggg!!!

[midnight pier]

Thank you friend

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[cursive ridge]

Hello, my lecturer give me a method for this exercise but i quite don’t get it. Can anyone help

[copper stump]

You got A as
$ \begin{pmatrix}
2020 & 1 & -2020 \
2019 & 2 & -2020 \
2019 & 1 & -2019
\end{pmatrix} $

[rocky lotus]

Βαχτερ10Φρ4γ

[copper stump]

and you need to find B according to the hint, can you do that?

[cursive ridge]

B= A - I

Is it?

[copper stump]

indeed

[cursive ridge]

So what will I do next

[copper stump]

your teacher also told you B^2 or B^3 = 0

you can try to find what exactly is the B matrix, and what powers of B are all 0

and replace the values in the expression you have to find

[cursive ridge]

Hm how can i find it

[copper stump]

just calculate this???

[cursive ridge]

What about B^2 and B^3

[copper stump]

did you find the B matrix?

nature of B directly tells about that

[cursive ridge]

Here it is

I just dont understand the connection between these

[copper stump]

now can you do B^2?

notice how all the rows are identical?

that would be helpful for the matrix multiplication

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@cursive ridge Has your question been resolved?

[cursive ridge]

So can i minus row to row so that i will get more 0 entries?

[copper stump]

no

I meant it make matrix product simpler

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[copper stump]

you can also notice that the sum of all the row elements (2019 + 1 + (-2020)) is 0

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[cursive ridge]

Thank u

I got it

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[midnight pier]

the formula to find the area of a circle is πr², but what do I get if I just use r²? i asked AI but it confused me more

[hardy coral]

... you get 1/π of the area of the circle

[wintry oxide]

You get the area of the square whose side is r ?

What are you trying to achieve ?

[midnight pier]

the circle is π times of square?

[outer belfry]

Yes, the Area of the circle is π times of the Area of the square

[hardy coral]

You also get the area of that smaller section of the circle, where the arc BC has length 2AB (in other words, angle BAC is 2 radians)

That area and the area of the square from @outer belfry are equal

[midnight pier]

π is the ratio of a circle's circumference to its diameter, how multiplying the square area to π gives the area of circle?

sorry for these stupid questions but i really want to understand it

[hardy coral]

You can find lots of proofs of that online; not sure anyone can explain it clearly without using a proof

[midnight pier]

ok thanks

[hardy coral]

Here's a simple argument for example:

The area of the regular polygon is OA times its perimeter, because each of these triangles have area OA times the side length divided by 2, and there are as many triangles in the polygon as the polygon has sides

As you increase the number of sides, OA approaches the radius of the circumscribed circle, and the perimeter approaches the circumference

So the area is simply the radius times the circumference divided by 2

-> r * 2πr / 2 = πr²

[outer belfry]

In our local primary school textbooks, they usually have similar "diagram" like these to "explain", non-rigorous of course:

[midnight pier]

i got it

thank u

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[neat tusk]

show steps of solving this binomial theorem problem

[pure mirage]

Do you see how to set up two equations?

[neat tusk]

no

[pure mirage]

Do you know the general formula for the binomial expansion?

[neat tusk]

yes

[pure mirage]

\li \ [(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}] i.e. for the first you have [(ax^3+b)^{8}=\sum_{k=0}^{8}\binom{8}{k}a^kx^{3k}b^{8-k}]

[rocky lotus]

Kepe

[pure mirage]

Do you see now what k you are looking at when x^(3k) = x^6?

[neat tusk]

i dont understand what steps ur taking

[pure mirage]

Do you agree with the picture I posted?

Just that those equations hold

[neat tusk]

i do

[pure mirage]

Alright. Well, now on the right side there is the binomial expansion of the left and we want to see what coefficient $x^6$ has. The right side looks like [(\text{some coefficient}_1) x^0 + (\text{some coefficient}_2) x^3 + (\text{some coefficient}_3) x^6 + \dots][ + (\text{some coefficient}_8) x^{24} ]

[rocky lotus]

Kepe

[pure mirage]

I.e. we just want to look at that one summand with x^6, right?

And ignore all the rest

[neat tusk]

yes

[pure mirage]

So we are looking for (some coefficient_3)

[neat tusk]

got it

[pure mirage]

The second sum here represents this addition of all the coefficient * x terms

We just need to find what k we should be looking at

Well, when is x^(3k) = x^6, for what k?

[neat tusk]

what

[pure mirage]

This thing

is just this in a more compact form:

[neat tusk]

what did you do

[pure mirage]

I wrote out (ax^3 + b)^8 using the formula for binomial expansion

[clever sentinel]

@neat tusk Do you know the binomial expansion?

[neat tusk]

i do but i dont understand why there's k's now

[clever sentinel]

Can you write it here or send an image? It will be good if you write the binomial expansion of (ax³+b)⁸

[neat tusk]

i cant write it

[clever sentinel]

you just need to write the first 2 terms and then the last term and use "..." between them

[clever sentinel]

Why?

[neat tusk]

i just know the formula i dont know how its applied

[pure mirage]

[this is what the notation with the Sigma (looking like an E) does in a short form; the k's are running from 0 to n, but if you hadn't had it then yes, write it as you know it]

[clever sentinel]

can u write the general formula you know

[marsh whale]

binomial expansion

[clever sentinel]

@neat tusk We can continue after you send the general binomial expansion formula which looks something like (a+b)ⁿ or (x+a)ⁿ. It should be present in your notes or in your book.

[vale dock]

@neat tusk Has your question been resolved?

[vale dock]

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[wet sundial]

need help checking Q from H/W

[hardy coral]

Looks correct

[stark wedge]

wait a minute

your work for Q1 looks cut-off or undone

[coral ocean]

There is a Euler circuit if and only if there are 0 or 2 odd degrees

[stark wedge]

G1 does not have an Euler circuit, because the degree of vertex E is odd integer and-
what next

[stark wedge]

0 only

2 allows an euler path but no circuit

[coral ocean]

what

[coral ocean]

oh okay

[wet sundial]

wait whats the diff betwwen path and circuit again?

[stark wedge]

whether you finish back where you started

[coral ocean]

and Euler circuit you meet every point only once?

[hardy coral]

No that's both

[coral ocean]

okay ty

[hardy coral]

A path doesn't necessarily have the same end points

(a circuit is a path)

[stark wedge]

thing is that if your graph has 2 odd-degree vertices then an euler path must start at one and end at the other

[wet sundial]

so its correct?

(other than hanging sentence)

[hardy coral]

"G does not have an Euler circuit because the degree of at least one of its vertices is odd"

If that's what you meant, then yes

[wet sundial]

ok thanks

.close

[vale dock]

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[opal pendant]

i found alpha = sqrt(2) by putting it in the curve

now im thinking i differentiate the curve twice? but im getting such an annoying expression

and im not able to do anything

[safe fulcrum]

unfortunately I don't see another way

[opal pendant]

😭

[safe fulcrum]

,w (2x + 2y * u) + 5(x^2 - y^2 - 1)^4 * (2x - 2y * u) = 0, solve for u

u = y' btw

sigh, I'll just use Symbolab to confirm your 1st derivative

yeah okay if the 1st derivative is still that impossible, hmmm

I think it's a product rule in disguise, yeah

[opal pendant]

hm?

[safe fulcrum]

if it were -3y^2 y' - y^3 y'', it would be easier

but unfortunately it's not

[opal pendant]

u didnt put x,y=sqrt(2)

[midnight iris]

differentiate wrt x?

[opal pendant]

what will that do

[safe fulcrum]

for the second derivative, you can't sub in numbers too early

[midnight iris]

thats what the que is asking?

[safe fulcrum]

you have to algebraically find the 2nd derivative and then sub in

[midnight iris]

its pretty easy just see x as constant

[opal pendant]

no i mean we can find y' by putting x,y=sqrt 2..and then differentiate are curve twice and we will get some y' terms which we can then put as the y' we found

[safe fulcrum]

apparently it's like this

[safe fulcrum]

yeah

[opal pendant]

is there any better way?

[safe fulcrum]

unfortunately, no

[opal pendant]

this dosent look humanly possible to calculate 😭

esp under time constraints

but i guess my q is clarified

thank u

[opal pendant]

how is the question asking that 😭

its asking 3y'

y' = dy/dx

why would u differentiate wrt y

and even if we do

why would we treat x as a constant

.close

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[cosmic lichen]

a,b,c are natural numbers

[cosmic lichen]

Suppose that
a²+3b²=2c²

Prove that b is divisible by 9

First I tried doing mod 3

So that
a²≡2c²

Every square number have either 0,1 mod 3
If a²≡1
1≡2
Is wrong
So a,c ≡0 [3]

So I wrote that
a=3k
c=3m

9k²+3b²=18m²

Devide by 3
3k²+b²=9m²

[opaque stratus]

are you guys math majors discussing these

[cosmic lichen]

Take mod 3 again i suppose

[fickle rose]

instead I'd note:
b^2=9m^2-3k^2=3(3m^2-k^2)

[cosmic lichen]

So b²≡0
So b≡0 [3]

[fickle rose]

oh wait I see, okay

[cosmic lichen]

Also work

[fickle rose]

also small nitpick 18/3=6, not 9

[cosmic lichen]

Right

Then again I suppose b=3n
So b²=9n²

[cosmic lichen]

Back to this

[cosmic lichen]

Or this

3k²+9n²=6m²

Then wat

[keen tundra]

divide by 3, you have the original equation back

[fickle rose]

note that that's your equation again, just divide by 3

[cosmic lichen]

Yeah

So k²+3n²=3m²?

[midnight iris]

try contradiction?

[cosmic lichen]

I don't think so

But maybe

[hidden terrace]

$2m^2$

[rocky lotus]

Miyagi

[cosmic lichen]

Yeah

[midnight iris]

why not

well then b/9 will be some p/q

[cosmic lichen]

K²+3n²=2m²

[midnight iris]

and then we can just prove that p/q is irrational

right?

[fickle rose]

well you've shown that we can write (a^2+3b^2=2c^2) as (k^2+3n^2=2m^2) after taking out a three factor from all terms

[rocky lotus]

PajamaMamaLlama

[fickle rose]

just repeat again

and you have divisiblity by 9

[cosmic lichen]

Can u show ?

[cosmic lichen]

Hmmm

[fickle rose]

I will do a rewrite to make it clearer

(k:=a_1, n:=b_1, m:=c_1)

[rocky lotus]

PajamaMamaLlama

[cosmic lichen]

Ok

[fickle rose]

we get from (a^2+3b^2=2c^2) to (a_1^2+3b_1^2=2c_1^2)

[rocky lotus]

PajamaMamaLlama

[fickle rose]

with (b=3b_1)

[rocky lotus]

PajamaMamaLlama

[fickle rose]

repeat the same argument as before to get (a_2^2+3b_2^2=2c_2^2)

[cosmic lichen]

Ohhh

[fickle rose]

then it must follow (b=3b_1=3(3b_2))

[rocky lotus]

PajamaMamaLlama

[cosmic lichen]

So just take mod 3 again?

[rocky lotus]

PajamaMamaLlama

[cosmic lichen]

Tysm

Another n question may I ask?

oke 1 min

[hidden terrace]

What happens if you repeat the process indefinitely?

[cosmic lichen]

x,y are non-negative integers.
Prove that we can write any non-negative integer as 5x+3y, except for these numbers (1, 2, 4, 7).

[cosmic lichen]

You'll get stuck after the second one from looks

[hidden terrace]

You can use the properties of the GCD I think

[cosmic lichen]

How

[hidden terrace]

The GCD of any two numbers x, y can be written as ax + by

[fickle rose]

you've constructed an infinite descent of integers so no natural satisfy the equation, kinda like the sqrt(p) irrationality proof

[cosmic lichen]

Yk the proof?

[hidden terrace]

Yes

[cosmic lichen]

What is it

[hidden terrace]

I'll link a video if you're interested

[cosmic lichen]

Okay

But then what

[hidden terrace]

https://youtu.be/VEDl5n2CGuo?si=GZJg5US2C9jeWtq_

[cosmic lichen]

Then wat

[hidden terrace]

I'll take a different approach actually

So it's obvious that 1, 2, 4, 7 are the only numbers under 10 that can't be written as 5x + 3y

We have 8 = 5x1 + 3x1

9 = 5x0 + 3x3

10 = 5x2 + 3x0

Also, if n can be written in that form, then n+3 can too

So by induction, all numbers greater than 8 satisfy the result

Does that make sense

[vale dock]

@cosmic lichen Has your question been resolved?

[cosmic lichen]

Can u repeat

[hidden terrace]

We know that 8, 9, 10 can be written as 5x+3y

And that n+3 = 5x + 3(y+1)

Which means that 8+3=11 satisfies the condition

So does 14, 17, and so on

Repeating for 9 and 10 gives the rest of the numbers above 8

Since 3 also satisfies the condition, only 1, 2, 4, 7 do not