help-13

[graceful karma]

So it both impacts steepness and frequency

[rough mango]

so in this picture, the range is 2 and -2 and also impacts its frequency?

[graceful karma]

The range of the derivative doubles

Sin of anything will always be -1 to 1 though

Anything real

[rough mango]

so I can say the wave is a bit squished since the derivative is twice as much change?

[graceful karma]

I suppose you could say its a bit horizontally squished

[rough mango]

hm, still confused how it affected its frequency

[graceful karma]

Cause angle is changing twice as fast

[rough mango]

Oh wait what

you mean if it is x, then the curve turns at 90, but when 2x, it turns in, let's say 45?

but it's till 90

[graceful karma]

Ye

[rough mango]

oh make sense

I thought it somehow added another wave, but it was just squished

thanks

.close

[cedar kiln]

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[graceful karma]

Np

You can look at the double angle formula too

It kinda is adding another wave

With a couple extra tricks

[cedar kiln]

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[modest vapor]

Can you guys check if my proof is alr?

[wraith dagger]

dulg

[modest vapor]

Are there anything wrong with this proof?

[daring geyser]

Answer please

[modest vapor]

ayo, i already occupied this channel first

[daring geyser]

Bro it's a humble request ..... answer please

[modest vapor]

man this gotta be against the rules: the help channel has my name on it

and it looks like you are cheating on an online test

and you haven't even showed your attempt, while i am here trying to verify my proof

[cedar kiln]

@modest vapor Has your question been resolved?

[agile bone]

that's not cool

[tropic oxide]

@daring geyser you need to go to an available channel.

also we do not give out answers

[cedar kiln]

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[crimson sedge]

is there a tutor here ?

[tropic oxide]

??

[crimson sedge]

Like they will do your homework for you and you gonna pay them something like that

[bright surge]

that's against the rules of this server

also don't reply ping ann

but don't post the same question on multiple channels

[tropic oxide]

you call that tutoring?

[dusky gate]

That's called cheating

[strong lantern]

send question you have doubts on, genuine good doubts though, not for sake of completing your questions try to think yourself first that's the point of solving maths

[cedar kiln]

@crimson sedge Has your question been resolved?

[sand ether]

[sand ether]

What even is the point of this? You will eventually have to sit actual exams anyway. If you don't wanna do your homework, then just don't do it. Like what's the worst that can happen? a detention?

[cedar kiln]

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[warm fable]

m kinda confused when to do-
nCr1 + n-r1Cr2 + n-r1-r2Cr3..... r_nCr_n
and when to do-
nCr1 * n-r1Cr2 * n-r1-r2Cr3..... r_nCr_n

[warm fable]

like there was a ques when we had 3 eqn and total six roots nd we had to find number of possiblities of their multiplication

nd someone said to add them is what makes sense but why

[sick ruin]

@warm fable send the context

[warm fable]

umm wait

why we add 15+6+1
shouldnt it be 15*6*1

this was my ques

@sick ruin ??

<@&286206848099549185>

[sick ruin]

Oh sorry

Was just trying to read your essay

Kidding

Ok hold on

[warm fable]

.

[sick ruin]

@warm fable why can you write r(x) as (x-a)(x-b)(x-b)

Like why is the b repeated

[warm fable]

thats typo

.

[sick ruin]

Oh it should be a,b,c?

[warm fable]

yea yea

[sick ruin]

Ok

@warm fable Did you come up with that? It’s a genuinely difficult problem and a really clever solution

Jeez

[warm fable]

its from some foreign olympiad

[sick ruin]

Makes sense

So what’s the issue exactly

[warm fable]

why we add those 3 cases in end?

in test i multiplied them

[warm fable]

thats y ask this ques

[sick ruin]

Oh yeah

Hm

That’s a good question let me think about that

[warm fable]

ping when u get it pls

[sick ruin]

Alright

@warm fable I‘m with you. I swear it totally seems like it should be multiplication. Once you pick the roots for Q(x)-1, you have 6 possibilities for the next two roots. That should mean it’s 15*6, not 15+6. It really makes no sense to me either.

[warm fable]

:)

[sick ruin]

Well do you have a teacher you can ask

[warm fable]

yea i texted him

[sick ruin]

Most of my teachers could not even read the question, let alone the answer

What class is this anyway lol

[warm fable]

junior year

highschool

[sick ruin]

This is standard stuff?

[warm fable]

or 11th grade

[warm fable]

no

m prepping for olympiads

[sick ruin]

Makes sense

What level of Olympiad

Are you in the US

[warm fable]

no

[sick ruin]

Mm

Well please DM me or ping me when you get an answer. I’m curious.

[warm fable]

ight

https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_24

@sick ruin i found another solution on AoPS

check it if u r interested

.close

[cedar kiln]

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[warm fable]

.reopen

[cedar kiln]

✅

[warm fable]

someone tell this please

[sick ruin]

Oh thanks, def still curious about the original tho

[cedar kiln]

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[cedar kiln]

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[nocturne kernel]

hi umm what is this symbol?

[woven path]

Normal distribution

[nocturne kernel]

kk thx

[cedar kiln]

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[dusky gate]

What have you tried so far?

[valid rampart]

Multiplying them

But I dont know what to do next

[livid hound]

make that alpha look less like a 2

consider starting with an equation using tan(a+b)

[valid rampart]

How?

[livid hound]

think about the hint for more than 1 minute

[valid rampart]

Ok Ill try

[valid rampart]

I solved it, thank you

.close

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[cedar kiln]

@left lily Has your question been resolved?

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[left lily]

NO

[muted tendon]

Looks like question is missing something.

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[crimson sedge]

Can someone help me with this? I don't know how to start on solving it.

[cedar kiln]

@crimson sedge Has your question been resolved?

[south juniper]

@crimson sedge For the first one, try to simplify right hand side and see if it comes out to x^3 - 2.

If it does, then it's an identity(since it's always true that x^3 - 2 = x^3 - 2)

For second one, you should know sin^2 (x) + cos^2 (x) = 1 is always true.

[crimson sedge]

what if it doesn't?

[south juniper]

then it won't always be true

[crimson sedge]

Oh, okay. How about the third one?

[south juniper]

try it urself first

[left lily]

The Cordouan lighthouse, located at the mouth of the Gironde estuary, measures 67.5m
high. A tourist at the top of the lighthouse watches the navigation and observes a sailboat as well
than a barge aligned with the base of the lighthouse.

1)How far from the base of the lighthouse is the
barge ?

2)How far is the sailboat from the barge

can you help me with this problem pls ? (i'm french)

[left lily]

sorry 😂

[muted tendon]

your pdf have given all the visual representation of problem , but the problem is i don't know french , sorry for that

[crimson sedge]

1. use trig func tan

2. use tan func again to determine the distance from the base to barge and then use tan function again to determine the distance from the base to the sailboat

distance = distanceOfbaseToBarge - distanceOfbaseToSailboat

thanks to google translate

[crimson sedge]

is it okay to square both sides?

[south juniper]

sure, then try to convert sin^2 into cos^2 or cos^2 into sin^2

Okay so (cosA - 1)^2 is not equal to cos^2 (A) - 1

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

[crimson sedge]

So,
sin² A = cos² A - 2 cos A + 1

[south juniper]

Yes

And

[crimson sedge]

What's next then

[south juniper]

sin^2 (x) + cos^2 (x) = 1 is true, so write it as cos^2 (x) = 1 - sin^2 (x)

Rearrange ur equation, you get cos^2 A - 2cosA + 1 - sin^2 A = 0

But 1 - sin^2 A = cos^2 A

So you get cos^2 A - 2cosA + cos^2 A = 0

Or 2 cos^2 A - 2 cos A = 0

Or 2cosA(cosA - 1) = 0

try to solve for A

basically you get cos(A) = 0 or cos(A) - 1 = 0. Solving for A gives you A = (2n+1)(pi/2) or A = (n)(2pi)

So the equation isn't true for all values of theta, it's only true for theta = (2n+1)(pi/2) or theta = (n)(2pi)

Hence it's not an identity.

I solved it for you cause I'm thinking you've got a lot of confusions.

[crimson sedge]

yeah, I really am confused. Thank you so much for answering!

[south juniper]

np but try to understand the solution

one other way is to look at the graphs of y=sin(theta) and y = cos(theta) - 1

you can immediately see that the graphs do not conincide, they only intesect at some points(points of intersection being the values of theta we found above)

[crimson sedge]

Hey @south juniper is this solution for the second one, right?

[left lily]

I translated into English just above the file

[muted tendon]

that first line

[crimson sedge]

Huh?

[muted tendon]

sin^2 A+ cos^2 A=1

ohh this question, my bad

[livid hound]

is way overcomplicating it

[crimson sedge]

I don't get it...

[muted tendon]

i think you can try these by putting random values , if for any value lhs not equal to rhs it will not a identity

[muted tendon]

this is real identity for sin^2x , cos^2x

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[thorny maple]

I’m stuck on number 2, I’m just having a hard time find the x-intercepts. I tried the diamond method and got -4 and 2 but don’t know what to do afterwards

[magic bough]

Ok so you found the roots?

@thorny maple

[obsidian coral]

Are you saying x = -4 and x = 2 or the factors are -4 and 2?

[magic bough]

Just find the maximum value

[thorny maple]

I’m say the Xs are -4 and 2

Saying*

[magic bough]

Because x has negative coefficient hence the graph is upside down "U"

[thorny maple]

Hold on, I will try to explain better

[magic bough]

You found when function intercepts x axis... Now find maximum value of function

[thorny maple]

That’s the work I’ve done so far but I’m just confused about if I’m right or not. So are my X intercepts -4,0 and 2,0 or am I wrong?

[magic bough]

x intercepts are good

Find max value of function

[thorny maple]

How would I do that?

Is there a formula or something

[magic bough]

complete the square

Because i am assuming you dont know calculus

2(-x2 -2x +8) this is your func right?

[thorny maple]

Yes that’s what I got

Wait

No, it’s -2(x2+2-8)

[magic bough]

I took the negativd sign in

[dense kayak]

you keep forgetting the middle term x btw

[magic bough]

^

[thorny maple]

What do you mean?

[dense kayak]

-2(x^2+2x-8)

[magic bough]

-2(x^2+2x-8) = 2(-x^2-2x+8)

Get it?

I took the negative sign in

[thorny maple]

-2(x^2+2x-8) was the one I got after I converted the original equation, which was -2x2-4x+16

[magic bough]

I know!!!

[magic bough]

This I did this

[thorny maple]

Nahhhh I get what you mean now

My bad dawg

Thanks for the help 💯

[magic bough]

You need to umm still find vertex

[crimson obsidian]

Vertex=maximum value?

[magic bough]

Vertex is at minimum or maximum value depending on function

Here it is at maximum

Because graph is an inverted U shape

[crimson obsidian]

Oh! Thanks m learning here

[magic bough]

Nice :)

[cedar kiln]

@thorny maple Has your question been resolved?

[thorny maple]

Yes

[woven path]

type ".close" to close the question

@thorny maple

[thorny maple]

.close

[cedar kiln]

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[frail dune]

which method do I use to solve it?

tried substituting $u=cos^{2}x$ but it leads me nowhere

[wraith dagger]

Erzis エルジス

[frail dune]

hm but wait, I get $du = -2cosxsinxdx = -2sin(2x)dx$

[wraith dagger]

Erzis エルジス

[crimson sedge]

One can write $\cos^2(x)$ in terms of $\cos(2x)$.

[wraith dagger]

Matthew8

[cedar kiln]

@frail dune Has your question been resolved?

[frail dune]

thank you

[cedar kiln]

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[crimson sedge]

Idk how to start this abstract alg proof

cancellation law

[wary nexus]

Or prove that a group has a unique identity element, then group => one element should be easy

One element => group is trivial

[crimson sedge]

can we start with =>

so that would be assuming that it is a group right?

which means its associative, has inverse, and identity

[wary nexus]

Yeah we need to prove
Group => one element (or contrapositive)
And
One element => Group (or contrapositive)

So something about the group-iness of the set will force all of the elements to be equal

[crimson sedge]

idk how to do that tho

like idk how the function plays a part in all of this

[wary nexus]

I.e. there is some condition that guarantees a unique element that will apply to every element of the group

It doesnt per se

It only matters that it forms group multiplication

You’ll see in advanced math, the literal definition of a thing doesnt matter as much as the properties it satisfies

[crimson sedge]

hmmm okay

so i have this one thing written in my notes

[wary nexus]

So * defines group multiplication, AND it has another special property (as a result of its definition) that basically forces all elements to be equal

[crimson sedge]

if ab=ac or ba = ca then b=c

corollary from that is

identity is unique, inverses are unique

[wary nexus]

Sure! So $\forall g,h \in G$ what property do we have that guarantees that $g=h$?

[wraith dagger]

ddxtanx

[wary nexus]

If we’re using cancellation law

[crimson sedge]

inverse

Actually, one only needs to show that all element $g\in G$ we must have $g = e$ where $e$ is the identity element.

[wraith dagger]

Matthew8

[wary nexus]

Yeah true

You can reduce it to that quite easily

[crimson sedge]

how would i start the proof formally

[wary nexus]

Well we want to show any 2 elements of G are equal

[crimson sedge]

cant i do proof bycontradiction

I did this earlier

Sort of, you can assume there is an element which is not the identity, and show that by direct proof it is the identity.

imma go to lecture rn

[wary nexus]

Yeah proof by contradiction is not necessary here

[crimson sedge]

do yall have any youtube recourses for modern alg

i think i just lack foundational knowledge

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[scarlet swan]

Hello, I have a question concerning this :

What's the trajectory ?

[elder parrot]

a vector having the components x and y i expect.

[scarlet swan]

but that's a : a segment containing the point O

a circle centered on O

a half line open at the origin

a segment not containing the point O

a line not passing through the origin

a circle not centered on O

I don't really know how to find the correct answer

<@&286206848099549185>

[cedar kiln]

@scarlet swan Has your question been resolved?

[cedar kiln]

@scarlet swan Has your question been resolved?

[cedar kiln]

@scarlet swan Has your question been resolved?

[cedar kiln]

@scarlet swan Has your question been resolved?

[gaunt hamlet]

The question says sin(ωt + π/2). If i recall, and correct me if I'm wrong, but to convert to cosine you need sin(ω(t + π/2))

Ah shit, you're right. I forgot that you'd actually have to shift by π/2ω, which will become π/2 after distributing. My bad.

[sturdy blade]

@gaunt hamlet Can u close if ur done

[gaunt hamlet]

Not my question

[sturdy blade]

Oh my

Bad*

[cedar kiln]

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[warm vector]

I’m having trouble in question 21

[floral thistle]

well

[sick ruin]

Think about it

You probably want to divide by r rather than multiply

[floral thistle]

dividing by a positive number less than one gives you a bigger number, doesn’t it?

[warm vector]

Ok

[golden sundial]

just give random values to p q and r

[warm vector]

Yes

[golden sundial]

yes what

[sick ruin]

Yes

[golden sundial]

let p be 2 and q 1

r 0,5

[warm vector]

But isn’t r 0

[golden sundial]

try each case

[floral thistle]

@graceful karma 🚰

[graceful karma]

Water well

[sick ruin]

They’re all positive

[golden sundial]

no

[sick ruin]

It might help to think about a concrete case; ie to plug in actual numbers for p, q, r

[livid hound]

wdym by

isn't r 0

[sick ruin]

It says r is a positive number less than 1

So it could be 1/2, 1/3 etc

[warm vector]

Oh right

Okay I got it

I wasn’t thinking thoroughly

[sick ruin]

Nice

[warm vector]

I have another question

[sick ruin]

Shoot

[warm vector]

Question 30.

[warm vector]

Lol

[floral thistle]

seems like you didn’t send the whole question

[sick ruin]

Well the number of eggs must be a whole number

[warm vector]

At the end it says collected?

[sick ruin]

Yes

The problem is fine

[floral thistle]

thought there could be some other valuable information in there oops

[sick ruin]

35% of n, which is 35/100 * n, must be a whole number

So

[warm vector]

So do I create an equation?

[sick ruin]

No

They’re asking what’s the smallest value of n where 7/20*n is still a whole number

[sick ruin]

Precisely

[warm vector]

Okay right I’m soo stupid

Thanks

[sick ruin]

Nah ur good

Np

[warm vector]

.close

[cedar kiln]

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[topaz rapids]

this might sound alittle simple but what exactly is it asking me 🙈

[topaz rapids]

is it asking for the average velocity ooor

do i just state the difference in height oor

do i state in words what the difference is

idk

[deft anvil]

Find Delta s

Delta s = s(15) - s(0)

I think you should get a negative number

[livid hound]

its asking for the change in height between the two specified times

to be more clear you can explicitly say that it increased/decreased by a certain amount (to be determined)

[deft anvil]

Delta s = final height - initial height

[topaz rapids]

so this right

@deft anvil@livid hound

[deft anvil]

What you found was the rate of change, or the average speed between those two points.

For your problem, you just to calculate the numerator of what you have in that picture

[topaz rapids]

minus 98.etc ?

[deft anvil]

It should be around -100

On the graph, the change in vertical position will be from the beginning point to where t=15.

Minus 98 or -98, yep

I found that by looking at the graph myself 200-100. Can you see how i did that?

[livid hound]

yeh. the height decreased by around 98 units

[topaz rapids]

idk why that was so hard for me to figure out lol

thanks guys

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[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[livid hound]

what have you tried?

well its a question about simultaneous equations

so the first step would be to set up your equations

hence

what have you tried?
was intended to see whether you did that

whut

you don't make much sense

(x+y+3) isn't an equation

neither is 3x-y+16

apply properties of the sides of a rectangle

and form actual equations

i.e. which sides will have equal length

uh huh...poorly worded. can you represent with some equations?

don't overthink this

is this a test>?

we can't help with tests

uh huh...

[silk sail]

It is against the server rules.

[livid hound]

not sure if this warrants a ping <@&268886789983436800>

[calm sierra]

ty

.close

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[lilac geyser]

how to solve this

<@&286206848099549185>

[vocal sparrow]

you can't prove that but you can prove that its square is 1

[lilac geyser]

how @vocal sparrow

[livid hound]

consider doing some squaring

[crude oriole]

Also try for taking lcm in 1st equation and substitute its value while doing 2nd one

[left lily]

tysmm 🙂 !

[lilac geyser]

ok

[cedar kiln]

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[crimson sedge]

hello i have a triangle with the following points: (0,8) (8,0) (0,0)
do you know how can i calculate its area?

[violet flume]

try drawing a picture

[crimson sedge]

sorry for bad english, am foreign

i did it

[violet flume]

what kind of triangle is it?

can you tell how long the sides are?

[crimson sedge]

uhhh

its like

wait ill send the pic

so i do 8x8:2?

[violet flume]

: is division?

[crimson sedge]

yes

[violet flume]

you can just use the normal formula, one half base times height

yup

its half a square

https://tenor.com/view/cat-cute-adorable-punch-gif-17822730

[crimson sedge]

how do i find height?

on general

[violet flume]

its 8

its how high it goes vertically

[crimson sedge]

k ty

[violet flume]

youll either need a line to go straight up

if you cant find that

you usually create a smaller right triangle

and use law of sines, pythagorean, or something

here you have it

[crimson sedge]

im 8th grade just had a blackout

[violet flume]

so you dont need to do anything special

[crimson sedge]

didnt learn that

yeah ty

[violet flume]

you will, no rush

[crimson sedge]

for the help

[violet flume]

np

[crimson sedge]

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[worldly basalt]

while practicing my algebra, I came up with this problem:

[worldly basalt]

$x^2 + 6x + 12$

[wraith dagger]

clockworkmurderer

[worldly basalt]

the idea being that I would find out which $(x-a)(x+b)$ could equal this

[wraith dagger]

clockworkmurderer

[worldly basalt]

however, it was soon clear that this doesn't factor evenly, and finding a and b such that 12 = a*b and 6 = a+b took my algebra skills past where I knew what to do next

[slim valley]

Are you willing to go to the dark side?

[worldly basalt]

my thinking is that I would do it with a system of equations, but it got really nasty really fast and I'm wondering what I would have to do in order to get past this. there's some algebraic "trick" that I can't think of that would get me through this (I think)

so what trick am I missing?

oh yes

yes I am

but if I don't get it, don't be mean to me. ;)

[slim valley]

Your function looks like thst

It never becomes 0

Use the square formula

Solve sqrt(-12) as sqrt(12)i

These are the two solutions

[worldly basalt]

that is vaguely familiar to me from high school but I've forgotten what the square formula is

[slim valley]

Top middle of the note

[worldly basalt]

ah

and am I seeing it correctly? the constant c is not under the radical sign?

hmm

it is underneath on the left side

[slim valley]

Your (x-a) (x-a) is

[worldly basalt]

got it; so my idea with the system of equations was doomed to failure from the start?

[slim valley]

No

It just doesnt have a real solution

[worldly basalt]

it felt like I was almost there but couldn't quite get the last few steps figured out

my algebra is so rusty and to be truthful I didn't really apply myself to learning it as a kid

[slim valley]

Wait a sec i messed up the - sign

Whoops

Thats the answer. I should really learn to double check

[slim valley]

Whats going on here?

[worldly basalt]

it's messy but; I wanted to use one of the identities I recently relearned to solve this:
x^2 + (a + b)x + ab

so given my formula, to find a and b, we have two formulas:

12 = a*b
6 = a+b

and going with my gut I subtracted b from both sides of the second formula in order to have a by itself.
then I substituted 6-b for a in the first formula to get 12 = (6-b)b and did some algebra to see if I could get b on one side by itself

[worldly basalt]

but I couldn't figure out where to go from there and soon started thinking that I was coming at it entirely incorrectly, as evidenced by your use of the square formula instead

[slim valley]

You can also simplify it

[slim valley]

You can use demovire theorem too i think

You cannot solve this like an algebraic expression

Say you had x^2-2x-4

[worldly basalt]

it's so odd that I was just practicing complex numbers yesterday and now I'm back here again XD

I haven't done the demoivre theorem yet though (I think)

[slim valley]

You can do that trick

[slim valley]

The essence of complex math

Look into it

Say you had x^3=27

We’d say x=3

[worldly basalt]

oh right ok I either forgot the name or the resource I was using didn't mention the name when it introduced the theorem

[slim valley]

But it has two more complex solutions

Using demovire its easy to find

Np

But make sure if you are making 2nd degree polynomials they cross the x axis once or twice

Or you have to deal with this

[worldly basalt]

🤔 what kind of 2nd degree polynomial that crosses the x axis once doesn't also cross it a second time?

[worldly basalt]

that's a very interesting comment though; I've never thought of it that way

but complex numbers are pretty new to me

[slim valley]

F(x)=x^2+x has one solution

But that would be at x=0 oops

Bad drawing

[worldly basalt]

by the way

is that recycled paper?

looks pretty coarse

so is that still considered an intersection?

I guess yeah that makes sense

as the limit at that point would be the same as the function value

😩 that's how I got started down this rabbit hole in the first place; trying to get derivatives down and I keep running into problems with my algebra skills

and realizing that I didn't remember how this factoring works

anyways thanks for your help!

.close

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[dense wing]

You only need 1 channel.

[strong lantern]

practice ?

revise concepts, solve textbooks, assignments

if you need more concept clarity you can solve solved questions

can you solve them without any help, or do you get stuck often

i mean there is not any krabby patty secret

just see what is hard

improve it

[dense wing]

If you're gonna use this channel, close your other channel.

[strong lantern]

solve more and more that type of questions

[crimson sedge]

Of course it is

Bye

Shut up

Bye bye cry in the corner

Thx mods

[limpid agate]

np

[cedar kiln]

@solid meadow Has your question been resolved?

[cedar kiln]

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[thin roost]

Yeah

[thin roost]

Ao

So

I'm learning

Calculus

And

First of all

How do you give like a lim symbol and a arrow

In latex

[south tundra]

$\lim_{x\to{a}}$

[wraith dagger]

Touch Our Beans

[thin roost]

Ok

$lim_{x\to2}(x+2)$

Huh

[south tundra]

\lim

[thin roost]

Oh

$\lim_{x\to2}(x+2)$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

So what does this

Mean

[south tundra]

This is the limit of x + 2 as x approaching 2, and saying that it equal to 4 basically means that

As x gets more and more close to the number 2

(x + 2) also gets closer and closer to 4

It says x -> 2

Whoops

Muscle memory lmao

[thin roost]

Ok this means when x arrives at 2 it can't go further

[south tundra]

When we're talking about limits, we usually don't talk about the case when x = 2

[modest vapor]

nono, it just means that along the function, you travel closer and closer to when x=2

[south tundra]

We rather look at the behaviour of a function of values near 2

[modest vapor]

the limit is basically "what does this function gets close to when our "x-variable" gets closer and closer to 2"

[thin roost]

Ok

So

[crimson sedge]

we dont talk about what happens at 2 but what happens as we get closer to it

[south tundra]

Yeah

[thin roost]

$\lim_{x\to\infty}(x+1)$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

So this means

It doesn't have a

Limit

Right

???

[south tundra]

Yes

It doesn't approach any particular value

It just goes upper and upper in the graph without stopping

[thin roost]

Oh I see

[modest vapor]

so you just say $\lim_{x \to \infty}{(x+1)} = \infty$

lol

[thin roost]

Oh

[wraith dagger]

dulg

[south tundra]

Yes, but we still say that the limit doesn't exist

Despite being able to write the "= infinity"

[thin roost]

Ok

[crimson sedge]

yes because inf is an idea - my calculus book

[modest vapor]

thats right

[thin roost]

What if

[modest vapor]

infinity is a concept of an random, large number

[thin roost]

$\lim_{x\to{y^2}}(x^2+z^2)$

[wraith dagger]

BASU乛ᴬᴳ

[south tundra]

Put the y^2 in {}

Ye

[thin roost]

Hm

Does smth like this exists

[south tundra]

Here you can just plug in x = y^2 basically, because the function x^2 is continuous everywhere

And you'll get y^4 + z^2

[thin roost]

So

[south tundra]

Yeah as long as z and y exist, this thing also exists

[thin roost]

This means

[south tundra]

But in the general case when x -> a, you can't just plug in x = a right away

[thin roost]

$\lim_{x\to{y^2}}(x^2+z^2) = y^4+z^2$

[wraith dagger]

BASU乛ᴬᴳ

[crimson sedge]

no no

y^2 + z^2

oh nvm

[thin roost]

But x holds 2

[crimson sedge]

i read it wrong

sorry

[thin roost]

Ok

Is my answer

Correct

Then

[crimson sedge]

yes it is

[thin roost]

Ok

Thx

.close

[cedar kiln]

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[crimson sedge]

ay I am gonna be in 9th grade soon like u

I got to ur website its nice

[cedar kiln]

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[charred sand]

does anyone know linear equation for inequality

[charred sand]

7x+8(x-3)=12-3x how to solve

[thin roost]

Distribute

8

$7x+8x-24=12-3x$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$15x-24=12-3x$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$15x=36-3x$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$5x=36$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$x=7.2$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$7x+8x-24=12-3x$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

Im wrong

@charred sand

So

Do like this

$18x=36$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$x= \frac{36}{18}$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$x=2$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

@charred sand
There you go

To check

Do

$14+16-24=12-6$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

$6=6$

[wraith dagger]

BASU乛ᴬᴳ

[thin roost]

Proved

[cedar kiln]

@charred sand Has your question been resolved?

[cedar kiln]

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[devout steeple]

Can someone please check this question on compactness for me

[cedar kiln]

@devout steeple Has your question been resolved?

[tropic oxide]

@devout steeple your picture for B is very wrong

but also, disregarding the bad picture,

you have established that it isn't closed

why go on to talk about boundedness when lack of closure alone already tells you your set is not compact?

(also the interior of B is empty)

[devout steeple]

okay, so B =!(1/n,0)x(0,1/n^2)

[tropic oxide]

(1/n, 0) × (0, 1/n^2) doesn't even make any sense tbh

[devout steeple]

whqat im confused about is B simply a coordinate or is it two intervals defined by 1/n and 1/n^2

[tropic oxide]

what do you mean by "simply a coordinate"

B is a set of points indexed by the natural numbers

[devout steeple]

okay now i understand, i was confused as i was thinking of it as if B={(x,y)| x is the domain of 1/n amd y is the domain of 1/n^2}

[tropic oxide]

whats this shit about domains now lmao

[devout steeple]

never mind, how would i then graph B

[tropic oxide]

you might notice that every point in B lies on the curve y = x^2

[devout steeple]

ahh okay, makes sense now what i did was so wrong

@tropic oxide what did you use to graph it?

[tropic oxide]

desmos

[devout steeple]

what did you write in desmos to get it?

@tropic oxide

[tropic oxide]

L = [1...100]
(1/L, 1/L^2)

[devout steeple]

thanks

.close

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[rugged tusk]

can someone help me to prove inequality between $\abs{arcsin x - arcsin y}$ and $\abs{x-y}$

[wraith dagger]

Michal

[rugged tusk]

using mean value theorem

[graceful karma]

Cant you just plug in a point

[rugged tusk]

no

[graceful karma]

Or do you mean inequality for every point

[tropic oxide]

do you have access to calculus

[graceful karma]

I mean he mentioned mvt

[tropic oxide]

oh yes true

[rugged tusk]

,rotate

Ended up with this

[tropic oxide]

consider that $|\arcsin(x) - \arcsin(y)| = |\arcsin'(c)| |x-y|$ where $c$ lies between $x$ and $y$

[wraith dagger]

Ann

[tropic oxide]

this is basically the same as what you wrote just restated differently

[rugged tusk]

but how to achieve inequality from this point ?

@tropic oxide

[tropic oxide]

consider the behavior of the derivative of arcsin.

[rugged tusk]

,rotate

Hope it's correct

[cedar kiln]

@rugged tusk Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[boreal storm]

hello! Can anyone help with this please? ```Consider a company manufacturing light bulbs. Denote by p the population defective rate
of their bulbs. The company wants to test the hypothesis H0 : p = 4% against HA : p < 4%.
A sample of 100 bulbs is randomly drawn and a number of 5 defective bulbs is observed.
We assume that the probability of observing a defective bulb is the same for each bulb, and
is independent of the other bulbs. Let the significance level α = 5%, and use the sample
proportion to perform the statistical test statistic.
Which of the following statements is true:

1. The rejection region of this test is [0.0722, ∞).
2. The rejection region of this test is (−∞, 0.0078].
3. The probability that the true defective rate is in the interval [0.0722, ∞) is around
   95%.
4. The probability that the true defective rate is in the interval (−∞, 0.0078] is around
   95%.
   Answer: 2. The rejection region of this test is (−∞, 0.0078].``` if anyone can explain why it's 2, i'd super appreciate the help

[cedar kiln]

@boreal storm Has your question been resolved?

[boreal storm]

it's been over 15 minutes so i gotta ping <@&286206848099549185> now?
if just something to help me on the way i'd appreciate

[cedar kiln]

@boreal storm Has your question been resolved?

[sick ruin]

Consider a company manufacturing light bulbs. Denote by $p$ the population defective rate
of their bulbs. The company wants to test the hypothesis $H_0 : p = 4%$ against $H_A : p < 4%$.\

A sample of 100 bulbs is randomly drawn and a number of 5 defective bulbs is observed.\

We assume that the probability of observing a defective bulb is the same for each bulb, and
is independent of the other bulbs. Let the significance level $\alpha = 5%$, and use the sample
proportion to perform the statistical test statistic.\

Which of the following statements is true:\

1. The rejection region of this test is $[0.0722, \infty)$.\
2. The rejection region of this test is $(-\infty, 0.0078]$.\
3. The probability that the true defective rate is in the interval $[0.0722, \infty)$ is around
   95%.\
4. The probability that the true defective rate is in the interval $(-\infty, 0.0078]$ is around
   95%.\

Answer: 2.

[wraith dagger]

abs_0

[sick ruin]

@boreal storm I don’t assume most people here know how hypotheses work, or what a significance level is. Is there any chance you could turn this into a purely mathematical problem?

[cedar kiln]

@boreal storm Has your question been resolved?

[boreal storm]

i don't know any of this. like, at all. i'm fairly stupid.
i'm trying to help a friend who doesn't know, and everyone in his class doesn't know

[sick ruin]

Lolll

Ok I’ll try to look some of this up

[boreal storm]

any help you can offer would be incredibly appreciated

[sick ruin]

https://courses.cit.cornell.edu/econ620/reviewm8.pdf

Ok so I found this

It’s super dense though, I highly doubt this level of rigor is necessary

Ok so I just learned that $$\alpha = \mathrm{P}(\text{reject }H_0 \mid H_0\text{ is true})$$

[wraith dagger]

abs_0

[sick ruin]

This is called Type I Error

[boreal storm]

alright 🙂 thanks. i'll pass this on

@cedar kiln

[sick ruin]

Hm

[boreal storm]

how do i close

[sick ruin]

Still trying to see this lol

[boreal storm]

oh? i thought that was...

sorry. i'm dumb.

[sick ruin]

Post this in #probability-statistics @boreal storm

[boreal storm]

i have no math skills.

[sick ruin]

Lol you’re fine

[boreal storm]

alright. done.

[sick ruin]

@boreal storm post this version, it’s easier to read

Reading https://online.stat.psu.edu/stat500/lesson/6a/6a.4/6a.4.1 rn

Actually kinda makes sense

The critical value is 0.04

We reject H0 if the proportion of defective lightbulbs is > 0.04; this is called the rejection region

[cedar kiln]

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[pure moss]

Uhm hi Guys i ll translate it .
F differentiable function
lim as x tends to +- infinity is + infinity

a<1

show that f(a)> f(0) + 1

[foggy merlin]

dommage qu'on a pas une idée des variations de f

[pure moss]

Thought abt using finite increment theorem for the first one [a,0]

On peut pas utiliser le theoreme d accroissements finis ?

et vu qu on a limite en +l infini aux bornes de R cela montre que f change de variations non?

[foggy merlin]

oui f change de variation on le voit bien avec les limites

[pure moss]

Si j utilise le TAF

[foggy merlin]

utlise le theoreme entre a et 0

[pure moss]

j aurai f(a)-f(0)=f'(c)a

Oui

On sait que a<1

[foggy merlin]

oui

[pure moss]

le prob c est que je dois montrer que f'c est bien aussi negatif pour obtenir que f'c . a > 1

[foggy merlin]

On doit chercher ce que représente c entre a et 0

[pure moss]

Comment?

j ai pense a ecrire f'c autant qu une limite avec x->c

[foggy merlin]

normalement f'(c) cest la pente de la courbe si je dis pas de betise

[pure moss]

oui

la pente de la tangente

[foggy merlin]

si seulement on avait une idée de comment varie f sur R-\{0}

[pure moss]

Ms on doit utiliser les limites je pense

[foggy merlin]

mais je me dis que f decrois a un moment

[foggy merlin]

oui avec les limites on remarque bien que f décroit à un moment donnée

[pure moss]

Oui de -L infini a un point donne

oui

et elle croit entre 0 et + l infini

[foggy merlin]

pourquoi penses tu qu'elle commence à croitre en 0 ?

[pure moss]

Le choix d intervalle ?

a 0 , 0 b

[foggy merlin]

c'est ce que l'exercice dit mais ça reste à prouver

mais j'imagine qu'elle décroit donc f'(c) negatif

on a pas d'info autre que les limites

[pure moss]

Comment puis je le rediger mathematiquement

[foggy merlin]

et puisqu'elle décroit forcement quelque part entre a et 0, f'(c) est negatif

[pure moss]

Non on a pas d autre info

Il peut etre qu elle decroit puis elle croit

pourquoi forcement?

[foggy merlin]

bah déjà, tu vas commencer par dire : d'après le TAF, il existe a, e € R-* et tel que f(a)-f(0) = f'(e)a

[foggy merlin]

si on veut representer la fonction en se basant sur les limites, tu verras que quelque part, la fonction décroit avant de remonter a +oo quand x tend vers +oo

[pure moss]

Oui

[foggy merlin]

par exemple, x²

[pure moss]

ms il peut qu elle est comme ca

[foggy merlin]

oui

[pure moss]

elle croit à un moment

Donc pour montrer que f'c est negatif

comment dois je faire svp

[foggy merlin]

Comme je l'ai dit, on n'a pas assez d'information sur comment se comporte vraiment la fonction mais on a une idée avec les limites données. La fonction décroit bien sur ton esquisse sur un intervalle entre a et 0, donc si on veut rédiger on peut peut etre dire qu'il existe un réel e sur ]a, 0[ tel que, f(a)-f(0) = f'(e) a

tout ce qu'on sait, c'est que c existe, mais on n'a pas d'ample info pour préciser réellement où

[pure moss]

On suppose que f est decroissante ?

ms ça serait une supposition

[foggy merlin]

l'analyse réelle est bourrée de supposition

[pure moss]

[foggy merlin]

et ce n'est pas une réelle supposition, puisque tu le démontres avec les limites

[graceful karma]

pain

[foggy merlin]

tu pars de +oo pour arriver a la fin a +oo, la fonction décroit forcément quelque part

[pure moss]

Bon on dit que qlw soit epsilon x<alpha => f(x)<epsilon

[foggy merlin]

pk tu parles de epsilon là ?

[pure moss]

Suis debile la

ça fait une journee que je suis avec cet exercice

[pure moss]

[foggy merlin]

un exercice qu'on peut torcher en 1h lol

[pure moss]

HHHHHH adhd

[foggy merlin]

quand tu n'as pas la réponse à un exo, passe à autre chose xd

et reviens plus tard

[pure moss]

Ouais c est cr que je ferai ms celui là m a trop perturbe

[foggy merlin]

pourtant il est pas chaud ton exo

c'est un exo d'application du TAF en vrai

[pure moss]

Merci pour me consacrer du temps je l apprecie vraiment ^^

[foggy merlin]

j'ai supposé que t'étais en prépa

[pure moss]

Ah un prof de 71 ans il donne tjrs des exos chauds

[foggy merlin]

71 ans

ah oui qd meme

le mec il a transcendé les maths

[pure moss]

A son point de vue ce nest meme pas un vrai exercice

J essayerai apres

[foggy merlin]

en vrai, c'est rien

[pure moss]

Merci encore une fois ^^

[foggy merlin]

np

[cedar kiln]

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[left lily]

Des français ?😱😅

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[main belfry]

HOW

i tried multiplying

multiplying binomials

it got even more confusing

[upper abyss]

Don't. That's already factored.

[gaunt hamlet]

If two things multiply to get 0, what does that tell you about those two things?

[main belfry]

oh shit

[upper abyss]

So now you can split it into two parts

[main belfry]

@upper abyss 11pi/6

thats what i got

[upper abyss]

cos(x) = -√3/2, sin(x) = -1/2

[main belfry]

oh

theres more than one answer apparently

they want ones where sin(x) = -1/2

and ones where cosx is -√3/2

[upper abyss]

Ye

11π/6 is a solution

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[sleek dock]

How would I justify this? I have stated the given and I am pretty sure its proven by AA

[sick ruin]

Probably AAA

Are you allowed to use English or

Do you have to write the proof in two-column hieroglyphics

[sleek dock]

Two column format

Is all it says

[sick ruin]

Hm alright

[sleek dock]

And I am really confused

[sick ruin]

Yeah this isn’t exactly straightforward

I’m guessing you would have to extend all those line segments into actual lines, and then use the properties of transversals

@sleek dock

Not sure if you’re allowed to do that though

[sleek dock]

Yea. Dont think so

Its Similarity proofs if thats any help

[sick ruin]

https://www.numerade.com/ask/question/show-that-two-triangles-are-similar-if-corresponding-sides-are-parallel-77434/

This is the exact same thing

But a different diagram

Maybe this will help

Okay I’m pretty sure the only way is if you extend the lines

[sleek dock]

I need an account for that

[sick ruin]

Oh that’s stupid

Ok draw the triangles on your own paper with all the vertices labeled and whatnot

And then extend ED and AC into actual lines

It will get messy, but I believe this is the only way

[sleek dock]

Ok thank you

.close

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[nimble onyx]

i need help

[south tundra]

Clearly the first answer should be the guaranteed wage of 25 + tips, which would be 17*14

So 25 + 17*14

And in the second one we just generalise this by making 17 equal to some arbitrary number t

So that'd be 25 + 14t

[nimble onyx]

39

[south tundra]

What

[nimble onyx]

u said 25+14

:/

[south tundra]

25 + 17*14

Because the waiter waited 17tables

[nimble onyx]

ye

[south tundra]

And for each table he gets tips of 14$

[nimble onyx]

ok

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@nimble onyx Has your question been resolved?

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[waxen plank]

I was doing a physics question and came across this integral. Can anyone anyone help me in this

[astral crown]

didn't try it, but looks like u=1-cos(theta) works

[waxen plank]

sin^3(theta) is causing problem

For u-sub

[south tundra]

Write sin^3(theta) as sin(theta)(1 - cos^2(theta))

And sub u = cos(theta)

[astral crown]

yea

[waxen plank]

Ohh let me try

[waxen plank]

I still didn't got could u elaborate

[south tundra]

$\frac{\sin(\theta)^3}{\sqrt{1 - \cos(\theta)}} = \frac{\sin(\theta)\sin(\theta)^2}{\sqrt{1 - \cos(\theta)}} = \frac{\sin(\theta)(1 - \cos^2(\theta))}{\sqrt{1 - \cos(\theta)}}$

[wraith dagger]

Touch Our Beans

[waxen plank]

How to proceed this step?

[floral thistle]

is that really much better

wait a second

this works

$\int \frac{1-u^2}{\sqrt{1-u}}\dd{u}$

[south tundra]

Sub u = cos^2(theta)

[wraith dagger]

quantum

[south tundra]

Ye you'll have that

[floral thistle]

we can split this into two integrals

since it’s subtraction

[waxen plank]

Ohh got it

[floral thistle]

the one with 1 as the numerator is easy

for the second one

sub in 1-u = v

so 1 = v+u, u = 1-v

expand the new polynomial

boom

[south tundra]

Yeah, or you could view 1 - u^2 as (1 - u)(1 + u)

[waxen plank]

What should I sub u = costheta or u=cos^2 theta

[south tundra]

cos

[waxen plank]

Ok thanks guys i think I got it completely