help-38

[past cargo]

so by HL these 6 triangles are congruent

so we can show the side lengths are congruent

yeah this one doesn't have jumps

thanks anyway though @prime lynx

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[prime lynx]

What is HL though

[past cargo]

HL congruence

not drawn accurately haha cuz i just drew it now

all these 6 triangles are congruent cuz they both have red and green

[prime lynx]

Yeah

[past cargo]

so AM = MB = BN = NC = CT = AT thus AB = BC = AC and thus equilateral

[prime lynx]

That's pretty cool.

[past cargo]

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[forest niche]

so we have a square inside a triangle

[forest niche]

H is the height

we're asked to find a (sides of the square)

can we say that AD || EF as well?

[pseudo dawn]

You have to show it how

[forest niche]

well

h makes 90 degree angle witH EF and the base right

therefore they are parallel

if im not mistaken

this is how it works

right?

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@forest niche Has your question been resolved?

[forest niche]

<@&286206848099549185>

can we by any chance say that CK = a/2?

[wraith hinge]

TF is THAT

[wraith hinge]

Soccer???

Or Squid game playground

[trim jolt]

@forest niche Has your question been resolved?

[forest niche]

<@&286206848099549185>

I think if we somehow find EK

we can just write a ratio

🤷‍♂️ not sure if there's an easier way

[swift iris]

What are you trying to prove

[forest niche]

not prove

just trying to find a

sides of the square

[swift iris]

Like a constant?

[forest niche]

yea

well

not a constant

we don't have any numbers

[swift iris]

Right

[forest niche]

it's gotta be unknowns

[swift iris]

I'll give you worked sol if I manage to solve it

[forest niche]

thanks 👍

this is the answer, can't understand why tho

actually

this entire thing i painted in yellow is a + h

just gotta figure out the ratio

[swift iris]

Is it to scale?

[forest niche]

well, the drawing is accurate

to the problem

[swift iris]

Right

[forest niche]

this is the original one

with no modifications

h is the height

[swift iris]

The unknown is the side of the square right?

[forest niche]

yea

a is

[swift iris]

Wait what

[forest niche]

"a" = sides of the square

[swift iris]

Individual side length?

[forest niche]

uh

all of the sides are equal

[swift iris]

Could you repeat what you need to solve again

[forest niche]

so we need to find a

via similarity

sides of the square basically

[swift iris]

So we're trying to find another equality...?

So like a=something

[forest niche]

yea

[swift iris]

Oh right

Okay

[forest niche]

a(a + h) = ah kind of gives the ratio away

a/ah = a/(a + h)

but that doesn't make sense

yea im silly

not sure where'd they get the a + h from

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[forest niche]

<@&286206848099549185> i'm stuck 🥲

[forest niche]

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[sonic kernel]

Hello I am currently trying to figure out a photonics problem but am guessing that I could use the mathematics disc to help me

[sonic kernel]

I have a device that has the following transfer matrix:

assuming i have two inputs to the device, where I only use 1 of the inputs as shown in the next image i get the results shown in E_out1 and E_out2

the issue I am having is that when E_in is equal to statement below I get totally lost in the results

[zinc ginkgo]

Are you using j and i for the same thing?

[sonic kernel]

yeah that was my mistake

i and j are the same thing

[zinc ginkgo]

What do you get lost about

[sonic kernel]

so theres two ways to approach this the way I did it out in the images and one where I directly plug in cos(wt) into E1 resulting in jcos(wt)=-sin(wt)

I am not really sure which way is the proper way to carry out the calculation

[zinc ginkgo]

Unclear when you're supposed to be taking the real part of the result

[sonic kernel]

do you have some sort of electromagnetics background?

[zinc ginkgo]

So far all you've shown is just matrix multiplication

No

Go to #old-network for physics server

[sonic kernel]

oh okay thank you

[zinc ginkgo]

Or EE server

[sonic kernel]

thank you

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[bright quarry]

you’re using the formula for the maclaurin series

it doesn’t work that way

[polar dragon]

isnt that taylor series

[bright quarry]

no

[polar dragon]

its c means its centered at pi/2 no?

[bright quarry]

because the center isn’t 0 you can’t just substitute x - pi/2

[polar dragon]

so i gotta do the whole chart thing for that

is my c also my a value

[bright quarry]

or just realize that sin(x-pi/2) is -cosx

and you should know the taylor series for that

centered at zero

[polar dragon]

what about this one

how do i do 2nd part

[bright quarry]

how many more questions are on the quiz

[polar dragon]

like 3 more but its all different kinds of problem

its not a quiz its like homework savas thing

[bright quarry]

good save

[polar dragon]

i got a exam in 3 days though

bro i aint capping

i got a 99 percent in the class

it lets me turn in the thing like 15 times

each problem

that aint no quiz

[bright quarry]

https://tenor.com/view/cap-cap-check-gif-24989004

[polar dragon]

look let me full snip it

only reason i cant do this rn is cuz i skipped like last 3 test

cuz my chem exam was fucking me over

[bright quarry]

alright

[polar dragon]

how does error thing work

[polar dragon]

yo tutorial rq for findign 10^-4 error

<@&286206848099549185>

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[tepid zenith]

can someone help me with these questions

[tepid zenith]

i'm unsure how to do them

and also i keep on getting confused with the bearing part

<@&286206848099549185>

<@&286206848099549185>

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[tepid zenith]

can someone help please

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[cedar nacelle]

can someone confirm why the phase shift is pi/2?

is it because in my general soution cos is 0, so in the final form we have to include the pi/2 as the phase shift?

[simple haven]

@cedar nacelle sin(t) = cos(t-π/2)

It is simply rewriting the sin as a cos for whatever reason

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[prime crystal]

it says in the question (last line)

[simple haven]

Ah! Good eye

Thanks!

[cedar nacelle]

right, but for some problems, like this one, the phase angle differs

I know how to do most of it up until getting the phase angle, im not sure if it has to do with initial condtions or what

[simple haven]

Have you talked about phasors or complex amplitudes before?

[cedar nacelle]

i think so

let me check my book

[simple haven]

Use complex amplitudes to combine sinusoids

[cedar nacelle]

have not, phasors are talked about but not for this section

we combine oscillations with phasors

[simple haven]

Yes

Use that technique

You can combine them as long as they have the same frequency

[cedar nacelle]

gotcha

[cedar nacelle]

and by combining you mean sin and cos

[simple haven]

Yup

[cedar nacelle]

gotcha

this would require a graph then no?

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[marsh forum]

Prove $gcd(n,n+1)=1$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

So what was thinking, to start $1\leq \gcd(n,n+1) \leq n$

[solid kiln]

Mr bean is not $\R \setminus \Q$

[marsh forum]

We now attempt to prove if $amod(n)=0 \implies amod(n+1) \neq 0$ and vice versa

[solid kiln]

Mr bean is not $\R \setminus \Q$

[tribal pond]

imma just stop you there

we have trivially that gcd(x,y) = gcd(x-my,y)

[marsh forum]

ooh

so gcd(n,n+1) = gcd(1,n+1)=1

[tribal pond]

yep

[marsh forum]

I've got to recall the theorms I've proven in NT

everytime, I just start from the ground up

[tribal pond]

but ur that person

so it's all good

im sure you would have gotten the right answer anyway

[marsh forum]

Thanks!

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[toxic moss]

Could someone explain this step

[simple haven]

@toxic moss if you think of z1 and z2 as vectors, then the result from the difference is also a vector.

If you think about a vector as a line segment from the origin, then this line segment has a slope.

The next line is simply a calculation of that slope.

[toxic moss]

Oh okay

Ohhh right I got it now tysm

And the argument is the angle the line segment makes with the origin yeah?

[undone river]

Guys I need help with math

Can anyone help me?

[toxic moss]

Channels occupied buddy

[undone river]

Ohk

Where do I go then

[simple haven]

Yup

Well

Specifically with the positive x axis

[toxic moss]

Aight tysm :)))

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[heady wind]

can someone help, idk what the lowest coomon multiple of 105 and 170 is, i just dont get it

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[nimble stone]

find their prime factorisation

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[snow sonnet]

2. A ball is launched straight up into the air and returns back to the ground 14 seconds later.
   A. Determine the initial velocity of the ball.
   B. Calculate the maximum height attained by it.

do i divide 14 into 2 since its the total time?

[tardy hemlock]

By conservation of energy, the speed that it goes up with will be the same as the speed it lands with

[snow sonnet]

so, 7s will be the time?

[tardy hemlock]

v = a * t = g * t, now for t, yes, you plug in 7 because from the turning point it takes 7 seconds to land

[snow sonnet]

alright so if i got it right, 68.67m/s is the initial velocity, then 240.35m for the height

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[odd tapir]

do holes affect the range of a rational function?

[lean prawn]

Define what you mean by a hole

[hollow tangle]

https://tenor.com/view/wink-eye-wink-gif-12352742718877841322

[odd tapir]

like at a particular point the function does not exist

[lean prawn]

ok I will just ignore the fact that the question you're asking isn't really well-posed and I'll give you the answer you want to hear because I'm a human and I can understand what you're really asking

If you add a "hole" to a rational function, it could affect the range but it doesn't have to. Consider for instance the rational function f(x) = x. If you add a "hole" to x=0 (i.e. consider the function restricted to nonzero real numbers) then the range goes from R to R\{0} (i.e. the range changes!)

Consider now the rational function f(x) = x^2. If you add a "hole" to x=1 (i.e. you restrict the function to those real numbers which are not 1) then the range remains {x in R | x >= 0} which it was originally

[odd tapir]

ok wait im confused

for example

take this

if we were to graph this

there would be a whole at (3, 3.4)

so how would u state the range for this equation?

[lean prawn]

Are you supposed to do this using graphing calculators or by inspecting the function by hand?

[odd tapir]

by hand

heres what i did

so would the range be y is an element of all real numbers

[lean prawn]

ahh that's nice

well done

the vertical asymptotes tell you that the function is surjective so the range is the set of all real numbers

the point where the function is not defined (i.e. the "hole") doesn't affect it because that value is attained by the middle part

[odd tapir]

ahhhh

i see

but lets say there was a graph that didnt have a middle part

[lean prawn]

btw I get a slightly different looking graph when I graph it

[odd tapir]

like this for example

then would the range be affected

[odd tapir]

wait rlly

how come

[lean prawn]

oh oops I just made a typo 😂

yeah that's good

[lean prawn]

For this function the range would not be the set of all real numbers because -1 (or whatever that is) is never reached with any input

[odd tapir]

yeahhh

makes sense

[lean prawn]

so again, we find the same conclusion that I laid out here

the range might change but it doesn't have to

[odd tapir]

yeah i get it now

thank u so much

[lean prawn]

no worries ^^ it was a very good question despite my shitty attitude

[odd tapir]

no no that was a great answer

dwai

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[flat flower]

Consider the relation R in Z defined by a R b ⇔ a² + b² is even.

1. Is R an equivalence relation or an order relation? Explain.
2. If R is an equivalence relation, determine all equivalence classes.

[flat flower]

so i proved its an equivalence relation

but what abt the classes isnt it just this

,rotate

[trail ingot]

[flat flower]

ello

[trail ingot]

i can’t quite read the writing 😭 is this supposed to be the equivalence class of 2?

[flat flower]

of r

SORRY

Kr = { x e R | xRr }

theres a third question tho

3. If R is an order relation, determine all maximal and minimal elements. Note: a < b ⇔ aRb.

[trail ingot]

oh well it should be ${x\in \bZ: xRr}$

[solid kiln]

generating function courtesan

[flat flower]

so maybe its not even an equivalence relation after all

so if a relation is defined into another set you use that set

so in this case its R in Z

so we use Z

[trail ingot]

yea. think of it like the elements in Z that are related to r

[flat flower]

thats a lot of sos

okok

do u think its equivalence or order relation

[trail ingot]

oh idk i just assumed it was when you said it was an equivalence relation and didn’t think about it haha

let me see

[flat flower]

,rotate

[flat flower]

thats what i did

those capitals R ill change to Z too then

[trail ingot]

a^2 + b^2 is even is equivalent to a + b is even. so yea i believe it’s an equivalence relation

[flat flower]

i just wasnt able to see a way to prove a would be equal to b

but antisymmetry worked

sorry i mean symmetry worked

haha

[trail ingot]

1. a + a is even for any a
2. if a + b is even, so is b + a
3. if a + b and b + c are even, that means a and b have the same parity, and b and c have the same parity. so a and c have the same parity and a + c is even

that’s my proof R is an equivalence relation

not very detailed though

[trail ingot]

for question 2 it wants you to describe the equivalence classes simply

[trail ingot]

so like here this is what an arbitrary equivalence class looks like

but try to figure out what, say, the equivalence class of 0 is

[flat flower]

so like x²=2k

[trail ingot]

sure. it’s the set of integers x such that x^2 is even (i.e. x^2 = 2k for some integer k)

[flat flower]

what wouldve you done

[trail ingot]

so i haven’t really explained what the point of this is but i would start with writing $$R_0 = {x\in \bZ: xR0} = {x\in \bZ: x^2 = 2k \text{ for some } k\in\bZ} = \ldots$$

[solid kiln]

generating function courtesan

[flat flower]

so like {0,4,6,..}

how many classes do i give tho bc cant u just go to infinity

[trail ingot]

no not quite

[trail ingot]

this is just the even integers

[trail ingot]

right but here there are actually only 2

one of them is the even integers, the other is the odd integers

[flat flower]

why would i want the odd ones

isnt the relation only valid for even outcomes

[trail ingot]

well we have e.g. 1^2 + 3^2 = 10, so 1 is related to 3

[flat flower]

oki

[flat flower]

ok but what can i change to this

because if i fill in like sqrt(2*7) u wont have an x thats in Z

[trail ingot]

right, but xR0 is true for every even x. because if x is even, so is x^2. and xR0 just means x^2 + 0 is even

and similarly xR0 is false for every odd x

[flat flower]

so taking the sqrt is making it more complicated in this case

its better to look for x itself

[trail ingot]

yea i wouldn’t look at the condition like x = sqrt(2*k)

do you know stuff about even and odd integers like for any integer x, x is even if and only if x^2 is even

[flat flower]

yeah

i was just making it too complicated haha

[trail ingot]

loll it’s ok

[flat flower]

now i see what u mean bc for the equivalence class of 1 its all odd numbers

[trail ingot]

yea so there are just 2 classes. one contains all the even integers, the other all the odd integers

all the even integers are related to each other and all the odd integers are related to each other

[flat flower]

yeah i see it now

tysmmm

[trail ingot]

yay

[flat flower]

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[sterile bone]

Would this be correct?

[sterile bone]

Differentiation

[cosmic meadow]

(Cosx secx)^2 = cos^2x sec^2x

[sterile bone]

Yes

But that wouldn’t help me differentiate

This is for derivatives

Using chain rule

[cosmic meadow]

But yeah

It is correct

[heavy palm]

I would recommand you use something like Wolframalph to check something like this.

[cosmic meadow]

You can do like that

[sterile bone]

Wait wouldn’t cos and sec cancel out to be 1

[cosmic meadow]

Your solution is correct anyways @sterile bone

[sterile bone]

I’m curious about this

Would it be shorter to do it this way?

After you apply the exponent to both terms

[cosmic meadow]

No, probably same lenght

[sterile bone]

What would you do

[cosmic meadow]

(d/dx cos^2x) sec^2x + cos^2x(d/dx sec^2x)

[sterile bone]

Yeah but my main thing is how do you differentiate when it’s cos^2

I know cos is -sin

[cosmic meadow]

U would get 2cos(x)sec(x)^2 * (cos(x) * tan(x) - sin(x))

Do u mean how do u derivate cos^2x? Not sure if i understand your question

[sterile bone]

Yes

[cosmic meadow]

-2sinxcosx

[sterile bone]

Hmm

So using chain rule

[cosmic meadow]

Or -2cosxsinx if u see better like that

Aha

[sterile bone]

You being the exponent down

[cosmic meadow]

It is a product

[sterile bone]

And then you multiply by the inner derivative too

[cosmic meadow]

Cosxcosx

[sterile bone]

Yes, but basically what you have is you’re using the power rule to bring your exponent down to the left side as a constant multiple and then you’re multiplying by the inner function as it is and then you’re multiplying by the derivative of the inner function, which is minus sine

In case I appreciate it thank you

[cosmic meadow]

Power rule + chain rule

[sterile bone]

Yes

[cosmic meadow]

(f(x))^2

Exactly

[sterile bone]

Thank you

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[fickle ledge]

Here's what I've done

Not sure how to evaluable E(1/d^2+2016)

[heavy palm]

basically, that's how I think of it.

[fickle ledge]

holy

ok

[heavy palm]

as divisor are symmetric.

and you add them up, it's one.

the only exception (if exist) would be sqrt(2016), but 2016 is not a perfect square (is it? idk).

[fickle ledge]

no it's not

2^5x3^2x7

so EV = 1/2?

[heavy palm]

I think so.

[fickle ledge]

alright, thanks

I'll write out the solution rq

yep, it works

[fickle ledge]

what's the thought process for coming up with that

[heavy palm]

intuition? basically disivor has something to do with the symmetric.

[fickle ledge]

true

alright

I'll keep that in mind in the future

thanks

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[sacred ether]

I tried to solve this use Lagrange multipliers, basically I get λ = 1 + or - sqrt14, how do I know which one to use to actually get the furthest one?

[heavy palm]

wait wait, do we have to use largrange multiplier method?

[sacred ether]

yes

[heavy palm]

I think it's a sphare. And therefore, connect the point with original point.

the intersection of the line and the sphare are nearst and furthest.

[sacred ether]

but is there a way to tell which λ to use to get furthest point, except just testing

[heavy palm]

Also, Lagrange is solving for the "deriviate" is zero. if you get multiple answers, you need to substitude to check.

not sure about that, but I think you can search for that. Seems like interesting. I mean, maybe like 2nd deriviate some kind of thing?

[sacred ether]

I just tested by simply testing both negative and positive λ, negative is further so I will just choose that, thanks.

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[late narwhal]

I don’t really need help with a problem/equation persay but I’m retaking my math test tmr and I forgot everything need some advice

[nimble stone]

alright, whats the situation looking like?

[late narwhal]

So on Wednesday I took my test and got a 5/16 I’m able to retake the test but forgot everything,

We went over the work cuz most kids are retaking it but I forgot that aswell

[nimble stone]

how many topics is it covering?

[late narwhal]

Gimme 1 min I have the quiz review

This is the quiz

I did bad on

The one I’m retaking

[nimble stone]

careful with your name my friend

[late narwhal]

Thanks

[umbral dragon]

It’s always smarter to not connect your real name with your dc account etc

[nimble stone]

but at a glance if you have time i think its manageable for you to have a good run through of the stuff there
depending on how much time you have exactly?

[late narwhal]

It’s not

[late narwhal]

I have till 2nd period tmr

[nimble stone]

what time is it now

[late narwhal]

7:24

I got till about 9

[nimble stone]

pm?

[late narwhal]

Yes

It’s 7:25 pm

I got till 9pm

[nimble stone]

hm, alright
take a look at expanding and simplifying expressions
that should get you a fair amount of the marks if you can get it down

then try have a look at rearranging equations/inequalities

the first thing should be your priority, while the second thing would be the last step of some of the questions

[late narwhal]

also

[late narwhal]

the last question here what would i work on to solve that

[nimble stone]

that falls under rearranging equations

[late narwhal]

is that everything

jus wanna do good on the retake

[nimble stone]

you can roughly boil it down to those two things yeah

[late narwhal]

thanks for the help

[nimble stone]

nw, good luck

try not to stress too much, and make sure to sleep enough

[late narwhal]

alright

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[meager widget]

I'm not sure how this works... but isn't everything right?

nvm got it

constant

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[waxen patio]

how do I solve this one with chain rule?

I don't get how it becomes 3^(xlnx) times ln3

[round imp]

do you know the derivative of 3^x?

@waxen patio

[waxen patio]

I forget it, honestly

[round imp]

oh

its 3^x * ln 3

[waxen patio]

there's so many rules I've had to memorize

[waxen patio]

why is that?

[round imp]

calc is like that

[waxen patio]

I don't understand how it works like that

[round imp]

good question

[waxen patio]

should I just take it for how it is

[round imp]

oh

wait

you know d/dx e^x is e^x by definition right

[waxen patio]

yes

[round imp]

$(\frac{d}{dx}\left(\textcolor{#B14BA5}{3^{x}}\right))(\frac{d}{dx}\left(\textcolor{#B14BA5}{e^{\ln (3)x}}\right))$

[solid kiln]

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

[round imp]

rip anyway the two are equal

right

[waxen patio]

so anytime it's b^x, the derivative is b^x times ln(b)?

[round imp]

yeah

[waxen patio]

what about like b^2x

[round imp]

follow by chain rule

[round imp]

then its a constant

like 3^2x = 2*ln 3*3^x

[waxen patio]

ok so the constant just moves to the front of the ln part

that makes sense

[round imp]

mhm

[waxen patio]

so back to this

[round imp]

oh yes

[waxen patio]

that's even more confusing because the exponent is a product rule in of itself

and the whole thing is chain rule

[round imp]

hmm

wait i need to work out

[waxen patio]

as in go to the gym

[round imp]

no

work out the answer

anyway let u = xlnx

then chain rule

then chain rule again

lol

[waxen patio]

where are you pulling a u from

[round imp]

u = x ln x

[waxen patio]

oh right

never mind

been a minute lmao

[round imp]

lmao

yea thats it

!noans remember

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[waxen patio]

so 3^u times ln3

I might be cooked

again

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[vocal hazel]

how does (+) imply that if x + y sum to an odd number than x and y have opposite parity?

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[tough moth]

Ah

[tough moth]

I return...

Step by step? O.o

[graceful steeple]

lowk

you cooked 💀

[tough moth]

Blegg

[spring finch]

first part you can simplify the 16/12

they both have a 4 in common so it becomes 4/3

[ruby cobalt]

cancel out the variables by subtracting the exponent and keeping the remainder wherever the bigger number was

q^2 on top, q^1 on the bottom, 2-1 = 1 and since the larger exponent was on top you keep the q^1 there (q^1 can also just be written as q)

[tough moth]

I'm still stuck rip

[spring finch]

what part

[tough moth]

Ah crap nvm-

I FORGOT TO SUBTRACT THE Q AND M'S LOL

Alright thank you two, ima leave this open til I finish assignment

[ruby cobalt]

okay 👍

[tough moth]

You're supposed to find 10 and 7's biggest denominator right?

[solemn sleet]

in the denominator, you could just factor out a -1

this will make it cancel with the numerator

[tough moth]

I guess 1 is the biggest

1 right?

Not a negative

Like that

[solemn sleet]

as in, -1 (10b -7)

[tough moth]

It was -1

Sadge

[ruby cobalt]

yea

bc if u do as dr pepper said

[tough moth]

Yeahhh

The answer was just -1

Idk how to get just the one

[ruby cobalt]

$\frac{10b-7}{-1(10b-7)}$

ah oops

[solid kiln]

Matz

[ruby cobalt]

there... omg that took too long

and that means u can cancel the 10b-7 on top with the 10b-7 on the bottom

which would leave you with 1/-1 and that simplifies to -1

[tough moth]

It was -10b/10b and -7/7

I thought that'd give two negative ones

[ruby cobalt]

you can't do that hello

you'd have to include the same denominator in both fractions

to separate it

[tough moth]

Can you cancel out a -7 on top and a positive 7 on the bottom?

[ruby cobalt]

yes

it'd give -1

[tough moth]

I really am dumb-

Alright

Lemme try

I got a new equation

Ggs thank you

Help with this one too? ;~;

[limpid belfry]

you divided by 2 wrong

[tough moth]

I did?

[limpid belfry]

2x^2 /-2x is not 2x

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@tough moth Has your question been resolved?

[tough moth]

So the answer would beee

That?

[tough moth]

Will come back tomorrow lel

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[flat flower]

how do u do this

how can i apply taylor rule

for sin(1/x)

[hallow wolf]

xsin(1/x)

use squeezee

[little glen]

You can also use LH rule

[flat flower]

do u mean l'hôpital

it said it doesnt work for it

[little glen]

yeah i just applied it

and it didn't work

@whole coral D:

[flat flower]

what do i apply squeeze on

[little glen]

Is the answer zero?

I think I solved it without using squeeze theorem

[flat flower]

yeah its 0

[whole coral]

You should hopefully know what x/sin(x) goes to, the suggestion was you find what x * sin(1/x) goes to as per here

[little glen]

You have to play around with the terms

[flat flower]

would squeeze even work?

bc wont you just get -x²<= x²sin(1/x) <= x²

and then u have 0/0 but it said l'hopital doesnt work

[hallow wolf]

x->0

[flat flower]

?

im confused

[little glen]

You apply squeeze on x^2 and - x^2

if you put zero in those functions, you get zero for both

and since -x^2 < f < x^2

as x -> 0

0 < lim x->0 f < 0

this implies lim x-> 0 f = 0

[flat flower]

yeah but that means x²sin(1/x) goes to 0

but then in the denominator u still have sin(x)

so 0/0

[little glen]

no that's where you have to be smart and split the limit into two :P

[flat flower]

yeah but thats bc lhopital doesnt work

[little glen]

apply squeeze only on the 2nd limit

[flat flower]

ah

ok tysm

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[little glen]

wlc

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[wraith hinge]

Is this correct?

[wraith hinge]

is this a test ?

[wraith hinge]

Khan academy

Checking my answer

its right

[wraith hinge]

thx

I knew it

!done

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[wraith hinge]

.close

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[wraith hinge]

Having a hard time solving this and I dont know where I went wrong
Im asked to solve this with the quadratic equation
x^2+4x+8=0

[spring summit]

u should get a complex number using quadratic equation

[wraith hinge]

ah

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[wraith hinge]

thx

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[stark drum]

I don't understand how these aren't equivalent

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[stark drum]

<@&286206848099549185>

[stark drum]

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[meager crown]

If (B \ C) ⊆ A and (C \ D) ⊆ A, then (B \ D) ⊆ A.
how to prove this,

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[lavish spear]

How do I simplify this polynomial

[zinc ginkgo]

Factor out a y^3 from two terms and y^5 from the other two

Then optionally factor out y^3 from all terms

[lavish spear]

I don’t really understand

Can you walk me through it?

[zinc ginkgo]

ab + ac = a(b+c)

Have you seen that before

[lavish spear]

Nope

[zinc ginkgo]

[lavish spear]

Sorry 😞

I’m really bad at math

[orchid wagon]

🤨

[zinc ginkgo]

https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_2e_(OpenStax)/07%3A_The_Properties_of_Real_Numbers/7.04%3A_Distributive_Property

Read the friends at the movies example

[lavish spear]

Oh

I know now

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[green dirge]

Help me do questiob 14, i already did a part

Sum of 1/9^i from 1 to inf is 1/8

[fringe gate]

hi pls help what is the derivative of f(x)= 1/(x^2-sec(2x^3-3))^2

plse help me

I keep getting it wrong

[green dirge]

I already occupied this channel

[fringe gate]

ok srry idk how this works

[velvet mica]

https://youtube.com/shorts/u0L7rvNumjw?si=laCg166YC86Kosm7 this doesn't necessarily answer the how to your question but it does give the answer

[green dirge]

its alr, go to an available channel then post it

[green dirge]

thx

[velvet mica]

It shows it visually just not algebraically

[bright quarry]

!help

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[green dirge]

how about sum of (1/8)^(j/3) run from 2 to inf?

[velvet mica]

I believe the j/3 is only exponentiating the inside sum

Wait nvm

Let me think

Ok i believe i understand now

[velvet mica]

So note that 1/8^(j/3) is the exact same as

8^(-3j), and you can immediately take 8^-3 to get (1/2)^j

Have you learned that the sum as n goes to infinity of (1/2)^n is equal to 1?

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[green dirge]

nah

btw another person alr help me, thx

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[velvet mica]

Oh cool sounds good

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[cloud vault]

how do i approximate for the value at c?

[nova spire]

you mean L(x) = f(a) + f'(a)(x-a)

a is your reference point

you're not gonna use c as your reference point since you don't know anything about it

[cloud vault]

oh my book used f(c) and stuff like that

[nova spire]

sure but it's only a notation

when it clashes with other notation, change it

[cloud vault]

okay then what would i use as a reference point

2?

because it’s close to 2.06?

[nova spire]

well it's also because it's the only point you do have information about f on

since it's the only reference point you can use, by default it's closest to 2.06

[cloud vault]

okay i’m gonna try plugging that in thanks

yeah it worked

thank you

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[primal bluff]

im doing intervals and sets, i need to do intersection of these two intervals and im pretty sure there is no answer to this, but i cant make sure since i dont have keys and we were learning this just today, can anyone confirm if my statement is correct and if not, how to calculate this? thank you

[fallen rivet]

Are your > brackets ] brackets? I’ve never seen someone use triangle brackets, but the first set of intervals dont intersect

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@primal bluff Has your question been resolved?

[primal bluff]

yeah theyre supposed to be closed brackets for some reason it just is like that in our math lessons. So how do I write that there is no asnwer to the first question, just a zero or is there a specific thing i should write?

[fallen rivet]

ø arguably this

but idk what your teacher wants or whatever

[thorn sparrow]

itll be either $\phi$ or just a plain old ${}$(representation of a null set/set with no elements)

[primal bluff]

alright ill write that, thank you

[solid kiln]

xd_senBugha

[fallen rivet]

yeah {} works too ig

[thorn sparrow]

the second set has intersecting intervals

[fallen rivet]

yeah

[thorn sparrow]

do you need help with that too?

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[primal bluff]

my bad, right now im stuck with the union of the first two intervals

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[primal bluff]

i thought the answer was (-2:2]

but the answers i tried to search upo just came out that its (-3:2]

.reopen

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✅

[primal bluff]

what i learned is that when there is an open half closed interval starting opened and ending closed as i wrote down, was that you dont write the number the opened interval starts on, you go to the next number

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[lime granite]

dude

atleast try

yes konohamaru ofc u do

[kindred pier]

!15m

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[lime granite]

dude dont help this kid too much

whih grade pancake

which grade

dude you dont know you are in what grade

where do u live america ?

where do u live

ok just the continent

you are such a brat kid

asian kid cant do these omg

[velvet mica]

Please do not say this to describe people

[lime granite]

I am also an asian yk

sure

ok

suppose you are in a car

ur dad takes 1 hr to drive 1mile

then how many miles will he drive in 2hrs

[noble bison]

Read the problem

What does it say

No

[fathom dagger]

No

[noble bison]

!nosols

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[noble bison]

I don’t care

Go to bed then

Do it tmrw

[fathom dagger]

Don’t waste peoples time

[noble bison]

We aren’t here to answer your questions

Maybe just read the the question

What speed does the Boeing travel

It’s right there

I can’t help but feel like you’re trolling

I wish you the best of luck

[fathom dagger]

It’s rage bait

[lime granite]

u know all abbrevation yapping and all the shitty things but cant solve problems of class 5

[zinc ginkgo]

<@&268886789983436800>

[zinc ginkgo]

just block and move on

[past imp]

yeah between the slurs and being rude against helpers I'm going for a mute here

.close

[trim jolt]

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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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[hybrid linden]

I got to cos^2θ = 3

Idk what to do next

[leaden atlas]

Where did you get that?

Because it is not possible (it would mean cos theta >1)

[hybrid linden]

ok so

5sinθ = 5(1-cos^2θ)

[leaden atlas]

No

There is no square on the sine

[hybrid linden]

oh

i did it wrong

so its sin^2θ = 1 - cos^2θ

ok wait

[stiff bluff]

Hmm

[leaden atlas]

Can you express cos^2 in terms of sin^2?

[hybrid linden]

1-sin^2θ

[stiff bluff]

Well express it like that

And rewrite ur equation

[hybrid linden]

yeah i know

[stiff bluff]

What will u get

[hybrid linden]

thanks for your help

.close

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Channel closed

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[wraith hinge]

damn

[trim jolt]

Available help channel!

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Remember:
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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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[terse portal]

hi

[terse portal]

i have a fast question

[wraith hinge]

hello

which part do you need help with

[terse portal]

Consider the parameter generated curve
a) Calculate the length of the curve. (2p)
b) When rotating the curve, a turn around the line x = −2 occurs
a surface of rotation. Calculate the area of ​​the surface of rotation.

on a

so i know which rule i need to use

but i dont understand which on i need to apply to dx and dt and dy

[wraith hinge]

well which rules have you found out here?

[terse portal]

this one

[wraith hinge]

great

[terse portal]

but the issue i have this this type of question is idk which part is which

[wraith hinge]

now to find dy/dt, you just differentiate y in terms of t and to find dx/dt, you differentiate x

y and x are both given in terms of t

[terse portal]

yes

and the thing idk the rule behind it

how to know which part is which

or rather i know dx/dt is for the x

and the other one is for y

[wraith hinge]

I am not sure I understand what you are trying to say

[terse portal]

so i wanna understand

what dx is = with

and so on

and how i need to think to get the correct answer

[wraith hinge]

hmm so you require help differentiating x and y eh?

I am not sure if I have interpreted your query correctly

[terse portal]

yea

[wraith hinge]

okay great

to differentiate y

[terse portal]

okay

[wraith hinge]

$$\frac{d}{dx} uv = u^\prime v + v^\prime u$$

[solid kiln]

Edmund Cloudsley

[pastel bay]

you guys doing a level maths?

[wraith hinge]

probably not

considering the test paper they have presented is not in english

but I suppose this should be A-level

[pastel bay]

the topic is the same

[wraith hinge]

yup

[pastel bay]

i just did this topic last week

[wraith hinge]

cool

this kinda math is usually taught in high school

anyways

[terse portal]

yea

i forgot this part

[wraith hinge]

no worries

[terse portal]

is half linear algebra

[wraith hinge]

remember u and v are two different functions

[terse portal]

and half of this xD

[pastel bay]

so theres product rule,Quitent rule.

[wraith hinge]

haha yes concepts in math are often intertwined

do let me know if you are able to differnetiate x and y properly

[pastel bay]

differentiation using sin,cos,ln,e

[terse portal]

so here

what would we get for dx/dt?

[pastel bay]

whats the question

[terse portal]

here

[pastel bay]

O_O its in a different language

[terse portal]

i just need to get the length of the curve

yea

bro my uni is so tarded xD

they have made half linear algebra and half of this

anyways i just need a reminder

[wraith hinge]

There’s not much linear algebra involved tbh 🤔🤔🤔

[terse portal]

of how to do dx/dt and dy/dt

[terse portal]

its 2 diffrent part

this is the second part

i focus on

so guys can u just explain fast to me lol how to do it for dx / dt ?

[wraith hinge]

Just use product rule

https://youtu.be/17X5g9QArTc?si=IIVDrHTAK52hxRPn

Watch this for a quick refresher

[terse portal]

yea got the answer

[wraith hinge]

Good

[terse portal]

.close

[trim jolt]

Channel closed

Closed by @terse portal

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drifting rivet]

If 4^4^5 = n^n, what is n

[drifting rivet]

a^b^c=a^(b^c) right

[golden wasp]

ye

[drifting rivet]

so should n=4^5

but n=4^4 beacuse 4^4^4^4 = 256^256 whihc is right

[bright junco]

well, a^b^c is not the same as (a^b)^c

the way you wrote it, a^b^c is just a^b^c, you cant do any better

[drifting rivet]

so how would i solve the question

is there a method i could use

[zinc ginkgo]

I don't even think the question is clear yet

Is the beginning
(a^b)^c or a^(b^c)

[drifting rivet]

the question is to solve for n

[zinc ginkgo]

The (4^4)^5?

[bright junco]

yeah so its a^(b^c)

[drifting rivet]

yeah thats the part im confused with is it a^(b^c) or (a^b)^c

[zinc ginkgo]

Ah okay that's clearer then

The former

,calc 4^5

[solid kiln]

Result:

1024

[zinc ginkgo]

,calc 4^1024

[solid kiln]

Result:

Infinity

[zinc ginkgo]

Gg

Write 4 and 1024 as powers of 2

[drifting rivet]

4=2^2
1024=2^10

[trim jolt]

@drifting rivet Has your question been resolved?

Channel closed

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Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gritty ether]

Having trouble with converting the following from polar to rectangular form: