help-10

[tepid pilot]

thanku for helping me

[mortal mason]

no worries !

[tepid pilot]

.close

[obtuse pebble]

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[obtuse pebble]

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[candid vessel]

i got it into polar coordinates and i got 2sin(theta)cos(theta) but i dont understand what it means by describe level curves

[viral blade]

Level curves are sets of solutions to f(x,y)=k for some k

Here ig you need k=0

[obtuse pebble]

@candid vessel Has your question been resolved?

[candid vessel]

i dont really understand

[obtuse pebble]

@candid vessel Has your question been resolved?

[obtuse pebble]

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[civic bone]

am i misunderstanding this

[civic bone]

or is (a) equivalent to asking for null space of the (1 -3 4 0) matrix

well, a basis of the null space

which would then be (3, 1, 0, 0) and (-4, 0, 1, 0)

but that is not correct

so where am i wrong

[shut lagoon]

d isn't bound by anything

So there is no reason this subspace would always have d=0

[obtuse pebble]

@civic bone Has your question been resolved?

[civic bone]

righttt

ok thanks

[obtuse pebble]

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[steel lagoon]

how do i know if this has
infinite solutions
or one solution

[viral blade]

well if you don't need to worry about the possibility of their being zero solutions then you can check the determinant

ignoring the bottom row

but it shouldn't be too hard to see the top equation will always give you exactly one solution for x_1

so just check the middle one to see how many solutions for x_2 there are

[steel lagoon]

what do you mean by checking the middle one

[viral blade]

the middle equation

[steel lagoon]

set it equal to rhs?

[viral blade]

well

look at the corresponding equation

[steel lagoon]

so should i calculate the determinant the 2x2

[viral blade]

no that's overkill and only works if you know there's no cases with zero solutions

which i realize you do not in fact know that now despite your initial message

[steel lagoon]

sorry im lost what did you mean by this then

[steel lagoon]

^

[viral blade]

you need to worry about the possibility of their being zero solutions

so checking determinant won't work

[steel lagoon]

should i set R2 = 0

and then plug the values into the right side to see if its consistent or not

[viral blade]

that would work

i'm not 100% confident you know why you're doing that

[steel lagoon]

youre right

i dont fully get it

im doing it kind of mechanically, if it works for other problems thats good

[viral blade]

i would write this out in equation form

probably makes it easier to visualize

[obtuse pebble]

@steel lagoon Has your question been resolved?

[steel lagoon]

i see, thanks

[obtuse pebble]

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[small shard]

why does this hold for ellipsis and circles? Its weird because the accumulated y on either side of this point x on a quarter ellipse is b/2. Why does an equilateral triangle appear?
https://www.desmos.com/calculator/ogeq0vnjev

[warm shale]

hikari

[small shard]

$$x = \frac{a \cdot \sqrt{3}}{2}, \quad y = \frac{b}{2}$$

[warm shale]

hikari

[small shard]

This was an error on my part, the next latex is correct.

[obtuse pebble]

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[radiant marsh]

I'm not sure if this is in the right direction

or do I do mod5 only to the final result

[kind hawk]

well the nice thing about modulo is that you can take mod whenever you want

for addition and multiplication

[radiant marsh]

do I have to do modulo only once in this example?

as in (a^5 + (b^5+c^5))mod5

[kind hawk]

whether you do a^5 + (b^5 + c^5 mod 5)^5 mod 5 or a^5+(b^5+c^5)^5 mod 5 doesnt matter

its the same result

[radiant marsh]

oh okay

so far I'm getting the impression that it is a field

do I have to prove all the 12 conditions one by one?

[kind hawk]

those conditions only involving multiplication clearly hold cause the multiplication isnt changed

but there is something different to notice here which will save you a lot of time

[radiant marsh]

is it about a + 0 = a

say a = 5, 5^5 + 0 mod 5 = 0, but a is 5

am I right?

[kind hawk]

well 5=0, so 5^5=0 is also clear

so not sure what you mean

[radiant marsh]

ah

[radiant marsh]

what is there to notice

I'm guessing it's related to one of the conditions

but they all seem to be true

[kind hawk]

how did you test them

just bruteforce check all options?

[radiant marsh]

I didn't test them all yet but bruteforcing seems like the only way

[kind hawk]

ok

[radiant marsh]

I don't wanna bruteforce

:(

[kind hawk]

have you tested the one for showing that zero is the additive identity yet?

[radiant marsh]

that's what I mentioned here though

[kind hawk]

but then you wrote something about a being 5

maybe you meant what I mean or not

what I mean is the following: a^5=a in Z_5

[radiant marsh]

what does that have to do with 0 being the additive identity

I'm a bit confused

[kind hawk]

thats the thing you have to check

that a^5+0^5=a

and 0^5 is clearly 0

so you just have to check a^5=a for every a

[radiant marsh]

shouldn't it be a^5 = a^5?

[kind hawk]

no

[radiant marsh]

so I should check if a^5 mod 5 = a for every a?

[kind hawk]

you have to show that $a+0=a$ for every $a\in\bZ_5$, aka $a^5+_50^5=a$ for every $a\in\bZ_5$

[warm shale]

Denascite

[radiant marsh]

yeah

0^5 + 0^5 mod5 = 0
1^5 + 0^5 mod 5 = 1
2^5 + 0^5 mod 5 = 2
3^5 + 0^5 mod 5 = 3
4^5 + 0^5 mod 5 = 4

[kind hawk]

good

so along the way we notice that a^5=a for every a

because the zero doesnt end up doing anything

[radiant marsh]

yeah

[kind hawk]

but that means the new addition is just the same as the old addition

you are adding a^5 and b^5 which are the same as a and b

[radiant marsh]

ohhhhhhhhhhhhhhhhhhhhhhhhhh

wait not oh

so we showed that this condition is true

but what does that help

does it mean that a^5 + b^5 is the same as a+b?

yeah it does

so it's (a+b)mod5

can I use that to prove that it's a field without proving each condition? Say something like it's a field because it's known that for (a+b)mod5 we already know that it's a field in Z_5

sorry if I'm wording it badly, I'm learning it in a different language

[kind hawk]

yes

its the exact same as the old one

as soon as you have shown that a^5=a for all a

[obtuse pebble]

@radiant marsh Has your question been resolved?

[radiant marsh]

ok so for my proof I proved that (a^5)mod5= a for all a in Z_5 by writing this and wrote that since (a^5)mod5 = a we get normal addition

[radiant marsh]

but how do I tie that into explaining why that makes Z_5 a field

[kind hawk]

like you just wrote

[radiant marsh]

saying that it's normal addition mod 5 is enough to prove that Z_5 is a field?

[kind hawk]

well presumably you proved before that Z_5 is a field with the normal addition and multiplication

[radiant marsh]

as in a previous question?

[kind hawk]

or in your course

in a lecture at some point

[radiant marsh]

maybe, I don't remember

[kind hawk]

look through your notes

[radiant marsh]

I'll gaslight my professor into thinking that we have proven it before

anyways thanks, I really didn't wanna bruteforce it and you saved me

.close

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[primal hatch]

Hello, for part B, are there any ways to solve this without trial and error? This is what I have currently where I'm plugging in b/a until I get an answer of 0 and the only method we've been taught as far as I'm aware

[full musk]

You could factor maybe?

[primal hatch]

How else could you find the factors tho?

The only reason I got the first factor was through trial and error

I have answer but I don't know how to do without trial and error for b)

Yes

It just wants to substitute that as an equation to solve for x

So part a) got theta = pi

Now just substitute theta as that equation so the whole equation = pi

That fits within the domain yeah

Yeah I got that part, I just don't know how to find factors without getting the factors of 252 over the factors of 12 and substituting them into the equation until it equals to 0

Yeah, I don't even know what she means by "concepts we've learned" cuz I even specifically asked her if there was any way to solve those types of equations during the lesson in case of a test and she said no

[obtuse pebble]

@primal hatch Has your question been resolved?

[primal hatch]

Just need another person to verify that there's actually no other way to solve it other than trial and error 🙏

[obtuse pebble]

@primal hatch Has your question been resolved?

[obtuse pebble]

@primal hatch Has your question been resolved?

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[obtuse pebble]

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[timid silo]

Can somebody do favour me in this?

[velvet rover]

what are you supposed to do?

[timid silo]

Can u explain me how they do solve this question?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Iss equation ko maximize krna h

Samajh gyiiin..

Range

Sunn mene yeh theta kisi value nikali then
Aayi

Theta aaya pie/4
And alpha is 3 pie/4 , 7pie/4 .. Is it sahi? If yk?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Nah, that's for horizontal ground

Here it's down inclined surface

[obtuse pebble]

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[slender acorn]

help pls

[slender acorn]

As Paťo walked through the door, he slammed his guitar against the doorjamb. Later on, when
he checked the damage, he noticed that one of the strings was tuned exactly a semitone higher
than the original frequency. Since the string was already due for replacement before that, Pato
figured he could kill two birds with one stone.
He decided to tightly and firmly wind a new layer of wire onto the string in the shape of
a helix. How thick of a wire, made from the same material, does he need to buy to retune
the string to its original frequency? The problematic string had a thickness of 1.1 mm at the
start. Assume equal temperament tuning and that the tension in the string remains unchanged
during this manipulation. Consider the string both in its original state and after the wire has
been wound, as a homogeneous circular right cylinder.

[high flax]

Could you help with question 30 and how to solve it

[tardy epoch]

,rccw

[obtuse pebble]

@slender acorn Has your question been resolved?

[obtuse pebble]

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[teal coral]

what is horizontal asymptote of f(x) = x-cos(x) / x

[short spire]

so youre assuming a horizontal asymptote exists or?

[teal coral]

well really its find the limit as x approaches infinity of that function

[short spire]

well, what have u tried?

[teal coral]

i tried saying that its 1 since the rate of x on top and bottom are equal

but when graphing it says that there's no horizontala asymptote

[short spire]

so youre saying the limit as x approches infinty is 1?

[teal coral]

yes

[short spire]

bcuz it looks like it might?

[short spire]

this reasoning is weak

[teal coral]

well what do you think

[short spire]

so since we're apporching x -> infty

then it's fine to consider when x > 0

now recall that cos(x) is bounded by -1 and 1, right?

[teal coral]

yes

squeeze theorem?

[short spire]

oops didnt notice u responded

[short spire]

no

now in particular -1 <= cos x, right?

[teal coral]

yes

[short spire]

then we have that -cos(x) >= 1

so since x > 0, then our expression x - cos(x)/x must be bounded below by x - 1/x

[teal coral]

yes

[short spire]

We can infact do better actually to really drive across the point that this does not have a limit

so do you recall that cos(x) <= x, when x large enough?

[teal coral]

wait it doesn't exist

[short spire]

yeah, why would it?

[teal coral]

because squeeze theorem states it does

[short spire]

nope

show ur reasoning

i can show that x - cos(x)/x is bounded below by x - 1 for x large enough

but x-1 is unbounded

so x - cos(x)/x approches infty as x goes to infty

[teal coral]

the limit of x-1/x is 1 and the limit of x+1/x is one

[short spire]

huh?

[teal coral]

so therefore x-cos(x)/x = 1

[short spire]

thats really bad

[teal coral]

.close

[obtuse pebble]

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[ebon lark]

Bullying the poor dude off of the channel

[short spire]

How is that bullying

[ebon lark]

" thats really bad "

Dudes clearly struggling

[short spire]

It was not meant to them but to the poor explanation

[ebon lark]

Either way it comes off as degrading

[short spire]

Im on the spectrum so i have a hard time seeing my tone trhough text (or others), nothing personal meant

sorry!

[turbid marlin]

ur completely fine, thats not bullying at all

you couldve been more clear and said "the explanation is really bad"

but otherwise nah u didnt do anything wrong

[obtuse pebble]

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[ebon zephyr]

im blanking out i understood how to get the answer to a

but now i have no clue how to use my calculator to plot the grap

i tried y= with the formula

[obtuse pebble]

@ebon zephyr Has your question been resolved?

[ebon zephyr]

<@&286206848099549185>

[ebon zephyr]

<@&286206848099549185>

.close

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[worldly bolt]

I need help with this

[viral blade]

Can you find tan(120) and tan(45)?

[worldly bolt]

Oh wrong question. It should be tan(80) instead of tan(120) and tan(20) instead of tan(45)

[rocky pelican]

this is tan compound angle

[worldly bolt]

I was trying to use sum and differences identities

[viral blade]

well that is the tan(a+b) formula with specific values plugged in

[worldly bolt]

So tan(100)

[viral blade]

right

[worldly bolt]

Oh ok for some reason it wasn’t letting me write it that way but now it’s working. Pretty sure it was just a technical issue

Thanks!

.close

[obtuse pebble]

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[viral blade]

Np

[obtuse pebble]

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[pale tusk]

Was there something off with the substitution, or did I make a mistake with sin(4x)?

[azure oriole]

sin is between 1 and minus 1

so its like solving ((1-4x+(22/33)*x^3)/x^5))

[plain stag]

1024/120 =/= 256/15

[azure oriole]

oh i didnt see infity but still same bs

answer would be infintiry i think

[plain stag]

you think incorrectly

their process is fine, it's just an arithmetic error

[pale tusk]

Man, I should really get some sleep

Thanks

[azure oriole]

my bad minus infity

nah bruh u wrong

[pale tusk]

It was 128/15

[azure oriole]

how

[plain stag]

marvellous

[azure oriole]

legit googled it

oh correct nm

nvm lmao

[pale tusk]

.close

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[acoustic reef]

43 I am getting -1/2

[acoustic reef]

Mb I got it

.close

[obtuse pebble]

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[vivid bone]

Why is Phi equal to the limit of the angle SQP as h --> 0?

[quick totem]

I guess it is because then Q-->P and that's exactly the definition of tangent line as a limiting position of the line QP.

[vivid bone]

I get that Q --> P, but I don't understand what you mean by the definition of a tangent line as the limiting position of the line QP

[obtuse pebble]

@vivid bone Has your question been resolved?

[obtuse pebble]

@vivid bone Has your question been resolved?

[vivid bone]

<@&286206848099549185>

[obtuse pebble]

@vivid bone Has your question been resolved?

[obtuse pebble]

@vivid bone Has your question been resolved?

[obtuse pebble]

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[simple pelican]

i dont know what integrals are

[dusk mirage]

Is this homework or something?

[simple pelican]

IXL

[arctic axle]

dawg if u dont know what integrals are

how can we help you

[dusk mirage]

Learn the topic first

Then ask questions or parts you don’t understand

[simple pelican]

well i have to do it so

[arctic axle]

lol

[dusk mirage]

Just learn it on khan academy or somewhere else first

[simple pelican]

what is that sign at the beginning and what do i do with it

[dusk mirage]

Its an integral sign, it means finding the area under the curve or finding the antiderivative

[simple pelican]

so would i take 1/2x and get rid of the exponent then do the top of the weird sign - the fraction then put in the bottom?

[arctic axle]

who is making you do it

your teacher?

[simple pelican]

my student teacher ;/

[dusk mirage]

You suould not be doing this without a sound understanding of integrals

Completely pointless

[simple pelican]

?

[arctic axle]

like

ur student teacher should not be assigning you calculus questions

if you don't know calculus

[simple pelican]

i just kinow power rule and derivitives

[arctic axle]

we cannot teach you this in less than an hour

[timid silo]

Maybe go on YouTube and search for a video regarding integrals

[simple pelican]

would this be right ntho

[dusk mirage]

No

[simple pelican]

oh

so khan academy is money

?

[arctic axle]

no

i dont think it costs anything

[timid silo]

No

Its free

Actually useful

Or u can also try watching the organic chemistry tutor on these calculus concepts i think he teaches them quite well

[simple pelican]

ill pay someone to teach me

like actually

also

im in 8th grade

[dusk mirage]

The internet is free

Asking for help here is free

No need to pay anyone

[simple pelican]

yes but idk how to

and i need taught

[sudden aspen]

hi

[simple pelican]

hi

[sudden aspen]

what’s up

[sudden aspen]

Integrals are also known as the anti derivative which can find the accumulation of change if applied

or areas

[simple pelican]

idk how to do that though

thats the problem

[sudden aspen]

you know what a derivative is, correct?

[simple pelican]

im in eigth grade

yes

[sudden aspen]

So what’s the derivative of ln(x)?

[simple pelican]

i can find like level 1 derivatives

[sudden aspen]

alright then we’re taking a too big of a step

let’s back up real quick

What math class are you currently taking right now?

[simple pelican]

acad pre calc

but im above my level

so my teacher gave to me

[sudden aspen]

I would say be very proficient in derivatives first before attempting to do integrals

[simple pelican]

i have to do it so

[sudden aspen]

you are required to learn integrals now?

[simple pelican]

not required but it was given

[dusk mirage]

Tell them you want to fully do differentiation first

It’s only understandable to introduce integrals at this point if the goal is to give brief overviews of topics but clearly as you re being asked specific calculation questions this isn’t the case

[sudden aspen]

Once you get to u substitution and beyond, it starts to become revolved around derivatives

If you want to learn calculus properly, I would suggest learning it in this order:

Limits
Derivatives
Applications of Derivatives
Integrals
Applications of Integrals
Differential Equations
Vector, Polar, Parametric Calculus
Infinite Series

[dusk mirage]

Make sure you know algebra and trig properly first, you need to be sound with logarithms, double angle rules, trig identities etc.

Trig is very important especially for integration

[modern pecan]

Know your inverse trig functions and hyperbolic functions

[dusk mirage]

I would say hyperbolics come in the middle of the course

[sudden aspen]

(so basically take precalculus)

[modern pecan]

Take ap calc ab

[dusk mirage]

exactly

[modern pecan]

For basic knowledge

[dusk mirage]

hyperbolics are pretty much never a calc prereq

[sudden aspen]

When do you ever use hyperbolics

[modern pecan]

I ask that question everyday

[arctic axle]

they show up sometimes

[modern pecan]

But it's trauma

[dusk mirage]

are you asking for real world applications?

[arctic axle]

$\sinh(x) = \frac{e^x - e^{-x}}{2}$

[warm shale]

rain

[arctic axle]

a very natural form to see

🙂

[modern pecan]

😭

Trauma

Also what did u do

[sudden aspen]

idek conic sections 💀

[dusk mirage]

If so then its used in activation functions in AI

[modern pecan]

How did you get that

[dusk mirage]

Thats how its defined

[arctic axle]

it's the definition 💀

[modern pecan]

Nooo

Ik the def

Like the function

How did you type kt

On discord

[dusk mirage]

tex

[arctic axle]

its called LaTeX

$a = 2$

[warm shale]

rain

[dusk mirage]

#latex-help #latex-testing

[sudden aspen]

is latex only used for discord

[dusk mirage]

No

latex is the standard for technical writing

[arctic axle]

latex is how professional scientists write their papers

[sudden aspen]

would that look good on a college application

[dusk mirage]

@simple pelican Can you close this

[arctic axle]

sure, but it'd be a small factor

[dusk mirage]

its off topic now

[dusk mirage]

eh not really

[arctic axle]

talk in discussion

[dusk mirage]

its just a touch at the end

[arctic axle]

yea

if you do math in college you'll learn latex

[modern pecan]

Interesting

[dusk mirage]

kinda pointless in HS though, Ive only used it for technical essays (which you dont do much of) and some proofs

[obtuse pebble]

@simple pelican Has your question been resolved?

[obtuse pebble]

@simple pelican Has your question been resolved?

[obtuse pebble]

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[late cradle]

Hi

[late cradle]

I’m really stumped on 7b and have no idea how to do it

Is the acceleration same for both objects

[dawn quail]

Calculate Ft

Wait hold on

[late cradle]

is there friction in this problem?

there is right

[dawn quail]

Well if it didn’t tell you don’t worry about it

[late cradle]

alr ill try do the problem disregarding friction

[modern pecan]

you have to use a matrix

[ancient drift]

Take the car and ball as a system

[modern pecan]

lemme show u an example problem

[ancient drift]

With the ball exerting a force of 500n on the system

[ancient drift]

So that is
a=500n/1050kg
=0.47m/s^2

Oh you got it

Nice

[modern pecan]

oh ok

[late cradle]

we learning the day before the test

[modern pecan]

i found an example problem

eh

guess u dont need it

[late cradle]

just send it

[modern pecan]

its solved

[late cradle]

ill do it for practice

oh

[modern pecan]

i was going to show

and point out

[late cradle]

[modern pecan]

of course for ur problem there was no friction

[late cradle]

ye

[modern pecan]

so force of friction wouldnt be there

but othere than that thats how u would solve an atwood problem

[late cradle]

whats different if there is frction?

[modern pecan]

u got a force going in the opposite direction of the acceleration

so now ur acceleration is different

you would need to use a sum of forces

[late cradle]

ahh ok

[modern pecan]

but if u had friction they would give you the coefficient of friction

so if there is no coefficient u can assume there is no friction

[late cradle]

got it

i was so confused

[modern pecan]

the car was in neutral

so free to move

its a different story if the car was in brake

[late cradle]

the word play is unreal in these questions

[modern pecan]

fr

[late cradle]

if im being honest he onlyever gave us one atwood practice problem

[modern pecan]

are u guys learning with friction?

[late cradle]

so i doubt he would put somehting like that on the test

[modern pecan]

if so i can send a bunch of my old problems

[late cradle]

ye we are

[modern pecan]

it gives u the answers so you can check your work

its not word problem

but it will give you practice with fbd

[late cradle]

These are the last type of problems I gotta review

[late cradle]

ok ty

[obtuse pebble]

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[shadow dagger]

Why is the function here increasing from (-inf.,3) and (0,inf) and decreasing (-3,0)

[shadow dagger]

like this example

The Y is going down as x increases so I would think it would be decreasing until -1.5

[opaque dome]

No first part decrease

It decrease till -3/2 and increase after

[shadow dagger]

says here increasing like this

[opaque dome]

Wth

[shadow dagger]

does it being a derivative effect it?

[opaque dome]

Oh this is f'

Well

if f' is positive then f increase

If f' negative then f decrease

[shadow dagger]

oh

[opaque dome]

Thats how they got it

[shadow dagger]

so different rules for a derivative?

[opaque dome]

Derivative gives monotonic

Of function

[opaque dome]

No deal with this

[shadow dagger]

I see

[opaque dome]

You compare derivative to x axis

[shadow dagger]

oh yeah that makes sense

well not actually

but I just needed the rule explained

thank you!

[opaque dome]

Here is the only thing you need

Youre welcome

[shadow dagger]

.close

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[wintry atlas]

Can I ask how I can get the value of x

[obtuse pebble]

@wintry atlas Has your question been resolved?

[sinful widget]

@wintry atlas Taking the square root of 336, which is 336 = 4(81)

x = 4sqrroot(21)

[obtuse pebble]

@wintry atlas Has your question been resolved?

[obtuse pebble]

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[viscid kelp]

How should I simplify this?

[obtuse pebble]

@viscid kelp Has your question been resolved?

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[novel siren]

can anyone explain why this is 10

[novel siren]

i got the vertical velocity is 15

from $\frac{2(v_{oy})}{g} = 3$

and in this setting $g = 10 \frac{m}{s^2}$

[warm shale]

aOsodP

[novel siren]

so if the vertical velocity is 15 m/s, shouldnt it be 15 at t=1

since d=vt

[warm shale]

aOsodP

[worn yoke]

could you explain where you got that formula from?

[novel siren]

its in the f=ma booklet

since it took 1 second to get to h1

and h1 and h2 are 1 second apart

it should take 1 second to land, giving 3 seconds of total flight time

[worn yoke]

ok so the 3 is the flight time, right

[novel siren]

yes

[worn yoke]

that's only true if there's no acceleration

[novel siren]

oh

so wait then the initial velocity, v0 would be 15

then using kinematics equation:
x = v0t + 1/2(-10)(t)^2

(15)(1) + (1/2)(-10)

15 - 5 = 10

[novel siren]

Oh okay that makes sense

so that would be true if we are talking about horizontal velocity

oh okay thank you!

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[fleet sierra]

I need to solve this non linear equation system the easiest and simplest way posible

[fleet sierra]

!status

[obtuse pebble]

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

[fleet sierra]

1

[polar fossil]

take the first equation, subtract the second equation

factor difference of squares

[fleet sierra]

Ok let me try that

[polar fossil]

actually don't factor difference of squares, you'll have a 5 on the right

but once you subtract the equations you should be able to do it with little problem

[fleet sierra]

After the substraction, the operation is supposed to be like this?:
X^2 = 9
(X-4)^2 = 4

[polar fossil]

no

,tex
\begin{align*}
\blue{x^2+y^2} &= \blue{9} \
\red{(x-4)^2+y^2}&=\red{4} \
\blue{x^2 + y^2} \red{- (x-4)^2 - y^2} &= \blue{9} \red{- 4}
\end{align*}

[fleet sierra]

what

[warm shale]

Хейли

[polar fossil]

we take the first eqn and subtract the second

this should be familiar from solving linear equations

[fleet sierra]

ok let me try now

i have a question

wait no

,help

[warm shale]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[fleet sierra]

the result of x should be 2.625?

[polar fossil]

dunno

does that work?

[fleet sierra]

rn im looking for the value of y to check

[obtuse pebble]

@fleet sierra Has your question been resolved?

[fleet sierra]

still on it

ok it isnt working y is 2.524 876 234 590 52 and when i resolve on the first equation it doesnt give me the 9

neither the 4 in the second one

what i did was with the linear equation expand the parentheses with binomial theorem

then resolve the equation as usal

at the end i divided 21/8 and it gave me 2.625

ok wait if i use the fractions it works

,close

.close

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[fleet sierra]

@polar fossiltysm

[polar fossil]

yes use fractions

and square roots and stuff

[obtuse pebble]

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[median summit]

i have to determine the domain, codomain, period, symmetry and sign of the following function y = 2sin(2x), in the interval [-π/2 ; π/2] i already determinated the domain but i dont know how to determinate the codomain

[median summit]

i have a test today im soo cooked help😭

[glad briar]

Ok so how do we approach the codomain of a function? For me it’s often helpful to visualize

@median summit

[median summit]

i think i would try to visualize the circumference

[median summit]

sry for late replay

[glad briar]

What values does sin(2x) take on the domain?

[median summit]

it would be all R but because of the restictions its [-π/2 ; π/2]

right?

[glad briar]

right that would be the domain of sin(2x) but I'm asking what values sin(2x) reaches inside that interval. Basically what is the codomain of sin(2x) here?

or if thats too much of a leap, what would the values of sin(x) be in that interval?

[median summit]

ehm bc of the fomdamentals of the sin it has to be in -1;1

[glad briar]

perfect

for sin(x) right?

[median summit]

yes

[glad briar]

So what happens when I multiply the x inside sin(x) by 2?

What does this change about the sin function

[median summit]

ehm it change smt on the domain?

[glad briar]

the domain is fixed here

[median summit]

i mean on the x axis

[glad briar]

we are always talking about -pi/2 to pi/2

what is the period of sin(x) without any changes?

[median summit]

2pi

i think

[glad briar]

yeah what about sin(2x)? 🙂

[median summit]

double of it?

[glad briar]

its actually the opposite

so the bigger I scale x by, the faster the sin wave is going to oscillate

think about what happens to x on the interval -pi to pi when I scale it by 2

[median summit]

its divided

like divise 2

[glad briar]

what do you mean by that?

[median summit]

wait

ehmm

[glad briar]

ok lets take a step back. First the question was what is the codomain of 2sin(2x) right?

We can talk about the period in a sec

[median summit]

okk thxx

[glad briar]

so you said that sin(2x) goes from -1 to 1 on that interval

[median summit]

yea

[glad briar]

can it be bigger than 1 or less than -1?

[median summit]

sin(2x) cant but if i moltiply everthing with 2 the interval doubles

[glad briar]

yeah so what is the codomain of 2sin(2x)?

you know sin(2x) has a codomain of [-1, 1]

[median summit]

-2;2 oh okk

[glad briar]

yeah perfect

do you need help with the period?

[median summit]

i can get it by my self i think but i have a question

[glad briar]

yeah shoot

[median summit]

to determinate the sign of that function i have to solve like 2sin(2x)>0 to see where its positive right?

[glad briar]

yeah that would give you the values of x where sin(2x) is positive

it can be hard to solve that in some cases but thats the idea

[median summit]

oh okk

let me try to see if i get the period

give me 2 mn

oh okk i got it

thxx a lot @glad briar

[glad briar]

great!

[median summit]

have a good day byee

.close

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[umbral furnace]

yo i need help with maths hw

[umbral furnace]

its

question 14

i alr used the substitution

but like

idk how to implement the Reimann Zeta function to the leftover part

some transformation

: (

[obtuse pebble]

@umbral furnace Has your question been resolved?

[umbral furnace]

<@&286206848099549185>

: (

@dapper bloom

ok imma just go sleep man

its like 3am

no one tryna help out

[obtuse pebble]

@umbral furnace Has your question been resolved?

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[burnt light]

there's 4 people. when you add their ages up 2 and 2, you get:
5, 11, 13, 14, 16, 22

[burnt light]

what is the age of the next oldest?

lets label them A, B, C and D, A being the oldest

A+B must be 22 and C+D must be 5

and so A+C must be 16 and B+D must be 11

ok so:

5=C+D
11=B+D
13=
14=
16=A+C
22=A+B

now there is A+D and B+C left

of which i am unsure of

however i am also unsure of if i need this information and if it is even obtainable

I can do ranges

so i am now lost

[arctic bloom]

when you add their ages up 2 and 2 ?? i dont really get this part

[obtuse pebble]

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[hollow cloak]

i kinda need help with a question:
prove that f(x)=x-sqrt(x^2+1) is 1-1, Df=R.
i cant do f(x1)=f(x2) neither do it with

[quick totem]

find derivative and show it is >0

[hollow cloak]

well the derivative is 1-(2x/2(sqrt(x^2+1))

and it is >0 only if x<0

[floral void]

Suppose f(x1)=f(x2)
x1 - sqrt(x1^2+1) = x2 - sqrt(x1^2+1)

(x1-x2) = sqrt(x1^2 + 1) - sqrt(x2^2+1) = (x1+x2)(x1-x2)/(sqrt+sqrt)

(x1-x2)(1-(x1+x2))/(sqrt+sqrt)=0

(x1-x2)(1-(x1+x2)) = 0

x1 = x2 or x1+x2 = 1

[quick totem]

if x<0 it is obviously positive

[hollow cloak]

hm lemme look at it

[quick totem]

if x>0 then x/sqrt(x^2+1)<x/sqrt(x^2)=1

so 1-x/sqrt(x^2+1)>1-1=0

[hollow cloak]

and how do we deny that x1 isnt x2-1

[floral void]

Just give a counter

x1 = 0 x2 = 1

[hollow cloak]

oh

thanks

[floral void]

Np ;p

[hollow cloak]

.close

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[quiet quartz]

How do I do the third one?

[clear cosmos]

Do you know where the a+b , an and a goes right?

Like how ab is c or -15 in your case

Nvm

I confused myself

[azure wraith]

What

So

a+b is the sum of roots

U can find that using -b/a

[quiet quartz]

Yeah I think ik what your talking about

[azure wraith]

I’m tripping

[quiet quartz]

💀

[azure wraith]

It’s gonna be

[clear cosmos]

AHH!!!

I remember

[azure wraith]

This

[clear cosmos]

Your supposed to remember couple of formulas for these question

[azure wraith]

And this

[clear cosmos]

And those equations will definitely use b.i and equate it to 16/4 or just 4

The quartic equation was like hell when proving

It goes along something like this

[clear cosmos]

For your question ^ not this

(a+b)^2 - 2ab

I think

Has to be it

Sub your values in

And equate it to 16/4
@quiet quartz

(-1/k)^2 - 2(-15/k). =16/4

Solve for k now

1/k2 +30/k.
= (1+30k)/k^2 =4

1+30k = 4k^2

It’s a quadratic

I’m sure you can solve it @quiet quartz

[quiet quartz]

This makes sense

[clear cosmos]

4k^2 -30k-1 =0

[clear cosmos]

What about the other lines?

Question solved then…

Cya

[quiet quartz]

Ty

[clear cosmos]

Np

[quiet quartz]

.close

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[cold brook]

can someone help me verify if my solution is right or worng?

[cold brook]

i am getting the divergence as

3/2 ( x^2 + y^2 + z^2 ) = 3/2(r^2)

in spherical

the volume element is equal to

r^2 sin theta dr dtheta d phi

and then i just solve this integral

[dark stirrup]

Since this is mentioning $\partial V$, does it want you to use Divergence theorem?

[warm shale]

SWR

[dark stirrup]

@cold brook yes. That's all correct.

I can't quite read your integral, but I'll trust it is correct since everything else you said was without error

[cold brook]

its basically r^4 sin theta dr dtheta d phi with limits being

0 to a, 0 to pi and 0 to 2pi. all of this multiplied with 3/2

[dark stirrup]

yeah that sounds right then

[cold brook]

cool thankyou so much

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[languid gulch]

The solution for the first part is wrong. Could they be asking in terms of x,y,s, and t?

[languid gulch]

this would be my solution in that case

I only have two attempts left

[floral void]

What

What is t’

[languid gulch]

OOPS

true

(2xy)(4t)+(x^2)(6t^2)

mbmb

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[lucid basin]

Suppose a robot moves from x,y coordinates (0.5 meter , 0.5 meter) to (0.65 meter, 0.7 meter). What would the angular rotation in radians be for each motor? Spool radius is 0.0045 meter. Base is 1.4 meter.

[obtuse pebble]

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[gilded hinge]

May I ask how to get the height of isosceles triangle PQS

Is PS the height of the triangle?

[cedar mantle]

can you make what we can assume clear here?

so are we assuming triangle PQS is isosceles?

and are we assuming triangle PQR is isosceles?

[tawny fog]

The question is very unclear and needs more information

Can't evaluate with just a figure

[gilded hinge]

It was given that PQR is isosceles, I have also figured out with the given in the question that PQ = PR = 39 and each side of SQR is also 39

This is the question In the diagram, PQR is isosceles with PQ = PR = 39, and SQR is equilateral with side length 39. Find the area of PQS.

I know that to get the area I need base times height / 2

But I'm not sure if the height is PS itself

[tawny fog]

Is SQR a equilateral triangle?

An*

[gilded hinge]

Yes

[tawny fog]

Find the area of SQR

You have the height and the base of Triangle PQR

[tawny fog]

Find the area of PQR from this

[timid silo]

PQR is equilateral

[tawny fog]

You can observe that triangle PSQ and PSR are congruent therefore
2PSQ(area)+PSR(area)=Area(SQR)
You have PSR and SQR you can do this by yourself

[timid silo]

This is not possible.

[woven wasp]

Hello, I need help creating an equation, So we have a webshop. We want to sell items on a marketplace. We want to have 7% profit on the sales from the overall price, but the marketplace takes 11% fee on sellings, from the overall price. How we can describe what price we should sell the items? Thanks for any help.

[tawny fog]

Isosceles

[gilded hinge]

May I ask why does it need to be 2PSQ?

[tawny fog]

#❓how-to-get-help

[woven wasp]

oh sorry

[timid silo]

SQR being equilateral implies that angle SQR = 60 deg
PQ = QR = PR = 39 implies that PQR is equilateral too.
Since angle PQR = angle SQR, the diagram is not correct and QS lies ON PQ.

[tawny fog]

See PSQ+PSR+SQR=PQR(in terms of area ofc)
And area of PSR and PSQ area equal (congruent)
Therefore
You can right 2PSQ

[tawny fog]

Yeah

@gilded hinge actually can you give me the statement of the problem

[tawny fog]

I think you have done something wrong cuz she is right

[timid silo]

!xy

[obtuse pebble]

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.

[gilded hinge]

Ok

[tawny fog]

Wth

[gilded hinge]

Only number 3

[tawny fog]

No like SQR is an equilateral triangle then SQ=QR=SR=PQ=PQ=QR and then the both triangles are equal 💀

And both the triangle coincide

Triangles*

[gilded hinge]

I'm sorry and that means..

[tawny fog]

The question is I'm pretty sure it's wrong

[gilded hinge]

Hmm ok

Oh wait

Now I get why it is wrong

[tawny fog]

Yep

[gilded hinge]

Ok thank you

[tawny fog]

Welcome

[gilded hinge]

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[stoic token]

.

[stoic token]

i have to check using SSS criteria if triangles are similar

I used the SSS criteria and got ratios as AC/PQ = 2/3, AB/PR = 2/3, BC/QR = 3/2

and this proves that the triangles are not similar please help me if I am correct

basically, I have doubt if I did it correctly

[cedar mantle]

um

i don't see anything wrong in your reasoning here

this shows that triangle ABC isn't similar to PRQ

however

if you want to show that triangle ABC isn't similar to PQR

then you would need to look at AB/PQ, AC/PR, and BC/QR instead

[stoic token]

y ?

[stoic token]

.

[cedar mantle]

ok so basically you currently are "assigning" AC to PQ, AB to PR, and BC to QR

technically, it is possible that a different "configuration" allows the triangles to be similar

where the sides in ABC correspond to different sides in PQR

[stoic token]

yeah, I understand what you are saying to me that there can be different configuration that can lead to the triangles being similar here but what I wanted to clarify is let's see if ratio is in my case as 3/2 which seems as if it is not properly turned that's why I had a little doubt

[stoic token]

.

can you tell me, please? What is the general rule of thumb to you Know assign corresponding sides to each other in triangles

like can you give sort of steps on how to check all possibilities?

[stoic token]

,

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[loud shell]

What theorem is this

[tardy epoch]

parseval or plancherel

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[inland dagger]

can anyone suggest me a resource which has problems about integrals

[inland dagger]

theoretical problems which are proof based

[pulsar terrace]

what kind of integrals

[cedar mantle]

spivak?

[pulsar terrace]

https://tutorial.math.lamar.edu/

[inland dagger]

for example something on the same flavour of

prove cauchy schwarz ineq in the integral form

[pulsar terrace]

chat gpt could prob help

i mean have u tried googling

what u need

[inland dagger]

this is like calc material i need problems for analysis

[inland dagger]

i tried but i thought someone has a great analysis II problem book

[inland dagger]

thanks i'll check that out

[cedar mantle]

oh for analysis

um highly recommend rudin

[inland dagger]

yeah but it's very hard to follow

and i need something which is problem pased

b

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[robust moon]

Second part of this question is writing it as a linear expression and I'm wondering how much farther down/simplified I need to go

[woeful edge]

r u saying 45 is the gcd of 1575 and 220?

[robust moon]

I'm saying 35 is, yes

[woeful edge]

35?

[robust moon]

yeah

.....

I got that wrong didn't I

[woeful edge]

uhhhhhhhhh

yes

u did

[robust moon]

it's 45 isn't it

[woeful edge]

i dont even understand ur reasoning to it

[woeful edge]

not even close

[robust moon]

.....

<w> Well help me be correct then?

[woeful edge]

gladly

do yk how the euclidean division goes

i meant

algorithm

[pulsar terrace]

gcd means greatest common denom?

[robust moon]

Barely

Yes

[woeful edge]

yes

[pulsar terrace]

is it not just 5

[robust moon]

No

[woeful edge]

yes

[robust moon]

what

[woeful edge]

lmao

ill help u get there dw

first, ur given 1575 and 220

which one is bigger?

[robust moon]

1575

[woeful edge]

good

so do the euclidean division of 1575 by 220

[robust moon]

1575 / 220 = Mod 7 R 35

[woeful edge]

uh

js say 1575 = 7×220 + 35

thats way better

[robust moon]

rodger

7 x 220 + 35

[woeful edge]

now were gonna keep doing euclidean divisions until we reach a remaining of 0

[woeful edge]

do yk the divisor and the remaining in this division?

[robust moon]

the divisor is the remainder, isn't it

[woeful edge]

no

divisor is what u divided by, remainder is usually the one with the "+"

in the expression

its what remains

after dividing a number

so if u consider the euclidean division of a by b:
a = bq+r
a is the dividend
b is the divisor
q is the quotient
r is the remainder

(note: r is greater or equal to 0 and stricly less than b, 0=<r<b)

if i confused u, say so

[robust moon]

lil bit

I do better with this stuff when I have the numbers directly correlated

[woeful edge]

ok lets say

we wanna do the euclidean division of 21 by 6

we get

21 = 6×3 + 3

yes?

[robust moon]

3 with a remainder of - ye

[woeful edge]

21 is the dividend

6 is the divisor

3 is the quotient

3 is the remainder

[robust moon]

okay

1575 is the dividend

[woeful edge]

yes