help-36

[tired walrus]

ok

!status

[final saddle]

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

[tired walrus]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tired walrus]

ok

first off list all the possible scores that a player can have @frail moat

and maybe also find their probabilities

correct. now work out the probability of each score.

looking at 1 player only rn

bro did you just use AI for this

!noai

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tired walrus]

i mean ok you have kinda shown that you don't want to understand anything. if you're just going to run to AI then what's the point.

ok then go back and work it out yourself and properly

make a table of all possible rolls of 2d6 and then in each cell write the player's score

and then write down the probability of scoring 0, scoring 1 etc up to 5

too early

we are still looking at only one player's score

the second player isn't in the picture yet

yes i want the distribution first

would like more separators than spaces between each score and its probability, but ok.

now make a table with P1's scores labeling the rows and P2's scores labeling the columns

paper exists

pen exists

photo exists

MS fucking Paint exists

if no effort then honestly i don't wanna help you anymore

i will just abandon you here bc effort

ok and who the fuck are you to tell me i can't say fuck

<@&268886789983436800> potential troll there

[steep hatch]

Swearing isn't against the rules, that'd be racist towards australians

well user decided to say a transphobic slur so I guess that handled itself!

[tired walrus]

lmfao what

i didnt even see it

[steep hatch]

automod

[tired walrus]

was it the 6 letter one starting with T

lmfaoooooooooo

[atomic moon]

How can you put the same react multiple times

[tired walrus]

get rekt loser

[tired walrus]

multiple servers

[hybrid adder]

getting mad in a math server

[atomic moon]

Ah i see

[tired walrus]

anyway

P1\P2 | 0 | 1 | 2 | 3 | 4 | 5 |
------+---+---+---+---+---+---+
    0 |   |   |   |   |   |   |
------+---+---+---+---+---+---+
    1 |   |   |   |   |   |   |
------+---+---+---+---+---+---+
    2 |   |   |   |   |   |   |
------+---+---+---+---+---+---+
    3 |   |   |   |   |   |   |
------+---+---+---+---+---+---+
    4 |   |   |   |   |   |   |
------+---+---+---+---+---+---+
    5 |   |   |   |   |   |   |
------+---+---+---+---+---+---+

copy this table down and then put a ❌ into each box which corresponds to one of the players winning

yeouch

i was thinking you would copy it on paper

or maybe you insist on doing everything in the discord textbox because the device you're talking on will self-destruct the moment discord is closed

ok anyway

there's too many crosses

i think you did exactly the opposite of what i asked for

the only outcomes in which there's a win are:
(0,4)
(0,5)
(1,5)
and inversions of these three

find the probability of each, knowing each player rolls and scores independently

then add them up

thats gonna be the probability there is a win

then take the complement to get the probability of a draw

[proper dagger]

[final saddle]

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[sharp wraith]

hi can i check that my understanding of the behaviour of the function f(x) = x + e^(1/x) when x < 0 is correct, the function is convex for -1/2<x<0 and concave for x<-1/2?

[final saddle]

@sharp wraith Has your question been resolved?

[final saddle]

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[near frost]

x,y,z are natural numbers(without 0) prove that sqrt(x^2+1)+sqrt(y^2+1)+sqrt(z^2+1) >= sqrt(6(x+y+z)) how do i prove this??

[leaden moon]

So you mean $\sum \sqrt{x^2 + 1} \geq \sqrt{6\sum x}$?

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

Easy one

[near frost]

yh whrre x's domain is x,y and z

[leaden moon]

Anyways

To solve this

[near frost]

how 😭

[leaden moon]

$x \in \mathbb N$ where $0 \notin \mathbb N$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

So what is $x \geq$?

[soft zealot]

1 divided by 0 equals Infinity

[near frost]

than 0

[leaden moon]

$\geq$

[soft zealot]

1 divided by 0 equals Infinity

[near frost]

idk wt you are saying

[leaden moon]

Wait a moment

I gotta spend so me time thinking about this

[near frost]

dw bro i spent 3 hours on this

i tried AM-Gm cauchy and titus lemma but i am new to competetive math so yh i am a noob

[leaden moon]

Ah oki

[tacit dust]

so we need to prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1} \ge \sqrt{6(x+y+z)}$?

[near frost]

exactly

[leaden moon]

Try letting the landing point of $x$, $y$, $z$ as $a$, $b$, $c$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

So this means $a + b + c \geq 3$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

For the sake of finding the landing point, we are just going to say $a + b + c = 3$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

@tacit dust

[soft zealot]

Thomas

[tacit dust]

?

[leaden moon]

You following?

[near frost]

yea kind of

but can you explain what a landing point is

[tacit dust]

the min of an equation

[near frost]

so the lower bound??

[leaden moon]

In AM-GM, the landing point is at $a_1 = a_2 = a_3 = \dots = a_n$

[soft zealot]

1 divided by 0 equals Infinity

[tacit dust]

well we are finding that so

[leaden moon]

Could also be the upper bound too

[leaden moon]

Now we need to apply AM-GM to specific terms

For instance, $\sqrt{x^2 + 1} + f(x, a) \geq \dots$

[near frost]

okei but what is f(x,a)??

[leaden moon]

Any expression in terms of $x$ and $a$

[soft zealot]

1 divided by 0 equals Infinity

[near frost]

okei i get it but what is a??

[leaden moon]

Should be the other way around or smth

[near frost]

AM > GM

idrk le me just give up

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

This one

[near frost]

but can you recommend me a book for competetive math

[leaden moon]

I don't have one

[tacit dust]

basics or what?

[leaden moon]

Now we want $f(a, x)$ to be something when AM-GM, the $x$ term should have a degree of $1$, as we want to turn the RHS into a form of $x + y + z$ along with smth

[near frost]

depends rlly i wn to learn for my countries Math Olympiad but i am low lvl rn like i barely passed the first phase + i am 11 grade but i wn prep for next year aka 12th grade

[soft zealot]

1 divided by 0 equals Infinity

[near frost]

thx bro but i dont understand

[tacit dust]

just asking, but which country/region are you in rn?

[near frost]

balkan country

[leaden moon]

Can you continue from here?

[leaden moon]

We want to have $x + y + z$ on our RHS

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

So for this one, we want something that when we AM-GM, we got something like $x$ on the RHS

[soft zealot]

1 divided by 0 equals Infinity

[tacit dust]

but x = y = z = 0 still fulfills in inequality, no?

[near frost]

no x,y and z cannot be 0 theyr natural numbers

dw js recommend me some books i should pick up??

[tacit dust]

Mathematics Olympiad Challenges has many great problems you can try to solve

[near frost]

this one?

alr ty thats all i need

how do i close the channel now

?

[tacit dust]

If you can use it to the full potential then it's rlly useful

[tacit dust]

type .close

[near frost]

thank you so much

.close

[final saddle]

Channel closed

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[final saddle]

Available help channel!

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Remember:
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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.

[winged rune]

Hey guys i have a calc exam tmrw and i am partially cooked please explain how do i solve a calc question that is
. X^2
f(x) = —————
. csc^4(2x)

[serene panther]

Wdym "solve"

You gave us a function

[winged rune]

I need the derivative

[serene panther]

Do you know chain rule

Or quotient rule for that matter

[winged rune]

Yes but when i get csc i get confused

Those trigonometry things

[serene panther]

You dont know the derivative of cosec x?

[serene panther]

💀 are you not familiar with trigonometry

[winged rune]

No

[serene panther]

💀

Buddy i suggest you start trigonometry before you enter calculus

[winged rune]

U come here to bully or help

[serene panther]

Im tryna help but you gotta learn some fundamentals

Do you know what sin cos tan is

[tired walrus]

aight hold on

0th thing

[winged rune]

Yes

[tired walrus]

. X^2
f(x) = —————
. csc^4(2x)

dont write things like this

f(x) = x^2 / csc^4(2x)

or, better yet, just send a picture.

[serene panther]

Cosec x is the reciprocal of sin x

[tired walrus]

now then,

[serene panther]

You know the derivative of sinx? @winged rune

[winged rune]

Yes

[serene panther]

Apply chain rule to find derivative of cosec x

And tell me what you get

[hybrid adder]

before that apply the product rule

[winged rune]

hold on let me write it down

[serene panther]

?

[wide swift]

there is no product rule

thats quotient

and youre supposed to do the chain rule first

in order to solve the quotient

[serene panther]

Alright lets not confuse the op let 1 person help

[hybrid adder]

bring the csc to the top by converting to sin

then product rule

I always stray from the quotient rule

[wide swift]

my b

[serene panther]

Wtf?

[winged rune]

Ki ik what u saying is just smth dumb 😂

[serene panther]

Alr guys i got this dont interfere

[serene panther]

You continue

[hybrid adder]

you really gonna solve this using quotient?

[serene panther]

... dude its the same thing.

[hybrid adder]

It's inefficient for this question

[winged rune]

Man if its weong dont bully me cuz i slept thru this lesson

[serene panther]

1. Thats (cosec x)^4
2. Its wrong

I told you only to find derivative of cosec x

[half raptor]

Bro if ya don’t get trig ur absolutely cooked in calc

[winged rune]

Our curriculum is smth else tbh ik

[half raptor]

That’s why precalc has trig and everything in it

[winged rune]

See we have this written in our notebooks

[serene panther]

Well then use it!

And yes, you have to remember this at some point

[serene panther]

Tbh if you practice, it'll automatically register in your mind

[serene panther]

Yes

[winged rune]

This is what is written

[serene panther]

Use that info

[winged rune]

I tried applying the csc(x)

[serene panther]

Are you confused in applying chain rule to find derivative of (cosecx)^4

[winged rune]

Hold on

[foggy ice]

i remember this by having two rows

row1: sec sec tan
row 2: csc -csc cot

if you want the deriv of sec, you go to row 1 and hide sec, whatever is left is the derivitive. so sec(x)tan(x)
if you the deriv of csc you go to row 2. and hide csc. whatever is left is the derivitive. so -csc(x)cot(x)

you just have to remember the two rows

you basically just hide whatever you want the deriv of, whatever is left is the deriv (in that row)

[winged rune]

How is it so far?

[winged rune]

Ok this seems like an easier way to find the derivative

Hold up

[serene panther]

This is perfect

[winged rune]

For csc^3 for chain

[serene panther]

Yeah okay

Now you can do quotient rule on the original question

[winged rune]

Ok continue on same path?

9&

Ok

[serene panther]

Yes

[hybrid adder]

@winged rune tell me if you want a faster method

I don't think the intended solution uses quotient rule either

[serene panther]

Bruh ok i understand now

What you meant previously

@hybrid adder can lead with it since i was acting stupid before

[winged rune]

Ok after i finish this so i can understand

[serene panther]

Nice

Thats good

You got it essentially

[winged rune]

Ok thank you for the help n for the initial bullying

@hybrid adder please show me ur method i want to try

[serene panther]

😭 but im being serious. You have to get familiar with trigo if you wanna do calculus

[winged rune]

Ud be surprised w our curriculum

Its our midterm exam tmrw btw

[serene panther]

💀 all the best i guess

[winged rune]

🥲

[hybrid adder]

okay

you knew how to apply the quptient rule right?

[winged rune]

Yes

[hybrid adder]

do you know how to apply the product rule similarly?

[winged rune]

Ik the product rule but im not sure how to apply it to this question

[hybrid adder]

do you have any idea how we could turn it into a product of two things?

there's a simple trigonometrix identity

the most basic one

[winged rune]

Huh

Bare w me

[hybrid adder]

all good

can you rewrite 1/csc(x) as another trigonometric function?

[winged rune]

It would be the reciprocal of csc

Ah yes i get u

But how will i apply produxt rule exactly?

[hybrid adder]

okay

no

so basically

the basic trigonometrix functions are sin(x), cos(x) and tan(x)

taking the reciprocal of each is csc(x), sec(x) and cot(x)

i.e sin(x) = 1/csc(x)

[winged rune]

Yes

Then come csc,cot, n sec

Yes

[hybrid adder]

yes

so what can 1/csc^4(2x) be?

[winged rune]

Take is as sin?

sin^4(2x)

[hybrid adder]

yes

sub that into the original equation and use the product rule to solve

can you do that?

[winged rune]

Hold up let me try

Wouldnt that be

x^2 sin^4 2x

[hybrid adder]

yes

[winged rune]

And from there just solve?

[hybrid adder]

yup

[winged rune]

Ah ok perfect

[hybrid adder]

much easier right

[winged rune]

Yes ok got u now

[hybrid adder]

a good rule of thumb is to always try to get out of quotient rule

I don't even have it memorised because it's so bad

[winged rune]

Cuz when i see how the q was i just went straight to quotient rule

[winged rune]

It works but i like ur more

Appreciate ur help man

[hybrid adder]

np

gl on the test

[winged rune]

Might come back here again cuz if i dont finish what i have within a span of 6 hrs i will fr be cooked

[final saddle]

@winged rune Has your question been resolved?

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[silk fjord]

I put in computer algebra system

the two big right traingles

the diameter

and the fact that the two big right triangles are similar

it spit out

$EH = 2\sqrt{5}$

[soft zealot]

UCYT5040

[silk fjord]

$FH=5\sqrt{5}$

[soft zealot]

UCYT5040

[silk fjord]

$AH=6 \ BH=15$

[soft zealot]

UCYT5040

[silk fjord]

So great now I plug these into that smallest right triangle, which is also similar

I get that OH=3

so then i use that in a giant right triangle

to find half of CD is $\frac{9\sqrt{5}}{2}$

[soft zealot]

UCYT5040

[silk fjord]

so then double is $9\sqrt{5}$

[soft zealot]

UCYT5040

[silk fjord]

and omg i just realized

haha

i got it right

forgot to do that

nvm

.close

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[leaden moon]

Translation: Given the equation $x^2 - 6x + m = 0$ with $m \in \mathbb R$ and is a parameter. Find all $m \in \mathbb Z$ such that the equation above has 2 distinct roots $x_1$, $x_2$ such that $2x_1^2 - x_1x_2 + 2x_2^2 < 38$

[leaden moon]

!status

[final saddle]

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

[leaden moon]

5

[soft zealot]

1 divided by 0 equals Infinity

[storm haven]

What is the work/solution you question?

[leaden moon]

wait a moment

@storm haven

[storm haven]

Which part exactly

[rocky tusk]

what’s confusing

[leaden moon]

all the work, i have to translate from my language to english

[leaden moon]

this is my work translated to english, can you check if it's correct?

[rocky tusk]

so it’s not someone else’s work/solution

[storm haven]

Seems all good to me

[rocky tusk]

4

[leaden moon]

oh mb

by the way, can you think what can i do with this question?

[rocky tusk]

nguyen

[storm haven]

Translate

[leaden moon]

Find all $m \in \mathbb N$ $(0 \notin \mathbb N)$ such that the equation $x^2 - (m + 3)x + m + 1 = 0$ has 2 distinct roots $x_1$ and $x_2$ which all different from 0 and $A = \left(\frac{x_1}{x_2} \right)^2 + \left(\frac{x_2}{x_1}\right)^2 \in \mathbb Z$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

yep

[rocky tusk]

have you tried anything?

[leaden moon]

im thinking of applying viete's again

[rocky tusk]

yea that would be reasonable

how can you express the sum of squares

[leaden moon]

hm

lemme think

$A = \frac{x_1^2}{x_2^2} + \frac{x_2^2}{x_1^2}$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

$A = \frac{x_1^4 + x_2^4}{x_1^2x_2^2}$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

or another approach is

[rocky tusk]

probably wouldn’t do that

ideally we want something with x_1 + x_2

and x_1x_2

[leaden moon]

if i let $t = \frac{x_1}{x_2}$ then $A = t^2 + \frac{1}{t^2} + 2 - 2$

[soft zealot]

1 divided by 0 equals Infinity

[rocky tusk]

yep

[frank shuttle]

It’s actually just nothing

[leaden moon]

i got $A = \left( \frac{(m + 3)^2}{m + 1} - 2 \right)^2 - 2$

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

that's oof

@rocky tusk

[leaden moon]

wym?\

[leaden moon]

you here?

[rocky tusk]

my bad i was doing my cwl attack

yea that looks good now what must be true if that’s an integer

also you should look at the other condition they gave you

ok i’m heading out

good luck

[final saddle]

@leaden moon Has your question been resolved?

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[nocturne lagoon]

Q2 and 3 i just cant solve, have no idea what im doing wrong

[storm haven]

They are not telling you to prove that
$$5 (2^k) = 10 (2^k - 1)$$
They want you to prove that
$$\sum_{i = 1}^k 5(2^i) = 10(2^k-1) $$

[soft zealot]

Matcha

[storm haven]

Remember that it is a sum

So, this is not the correct inductive assumption

[nocturne lagoon]

Ye i assume k is true im not proving that

[storm haven]

That's not what you should assume

[storm haven]

As I have said here

[nocturne lagoon]

Isnt that like what i did n=1 or is it different

[grand walrus]

I just got to know i can do this question in just 3 lines

[craggy plank]

Claim a new channel please

[slender oracle]

Hello sir I am having maths doubts

Please help

[final saddle]

@nocturne lagoon Has your question been resolved?

[shell condor]

wait just apply gp

what's the formula for definite gp, wait

[winter halo]

The topic is proof by induction

[shell condor]

Ahhhh, whoops

Yeah no way I'll be able to help then

[final saddle]

Channel closed

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[winter halo]

For n=1, there was only 1 term in the sum but when you assume it for n=k, you cannot assume there is only 1 term

[final saddle]

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[narrow blade]

how do i find the increasing , and decreasing for this problem?

[narrow blade]

i got stuck where i dont know how to find any x for the sin

x =0 , sin (3/2 (0)) = 0

is there any more x?

[nocturne lagoon]

Bring sin to the other sidr

Side

[narrow blade]

so i have to do inverse sin, to bring it over right ? @nocturne lagoon

[nocturne lagoon]

Ye

[nocturne lagoon]

Is there a range?

[narrow blade]

yes, 0 < x < 2pi

[nocturne lagoon]

Inclusive or not

[narrow blade]

no inclusive

[nocturne lagoon]

So not zero then

[narrow blade]

so, pi

[nocturne lagoon]

Where else in sin theta 0

[narrow blade]

oh, pi/2 , 3pi/2

[nocturne lagoon]

O mb it is 180

So 3/2 x=180

[narrow blade]

oh can i do radians

[nocturne lagoon]

Ye use pi if u want

[narrow blade]

i got thiss

[nocturne lagoon]

Simplify

[narrow blade]

simplify to pi/3

[nocturne lagoon]

Sub back in to check

Or look at the answer

[narrow blade]

i got sin(pi/2) = 1

i think i did something wrong here, sin(pi)

should be 0

[nocturne lagoon]

Its meant to be 2pi/3

[narrow blade]

yea

[chilly pollen]

@narrow blade draw the unit circle, it will be visually easier and then just be aware of the signs

find a general range and then change it to x terms

[narrow blade]

for the range is this correct?

[chilly pollen]

we need a range for the angle

@narrow blade like cosine decreases from $[0, \pi/2]$

[soft zealot]

Adhi

[chilly pollen]

I want you to find something like this for all quadrants

just simply $\cos \theta$

[soft zealot]

Adhi

[narrow blade]

cos theta,
decrease in Q2, Q3 ,
increase in Q1, Q4

[chilly pollen]

write it mathematically now

using ( ) and [ ]

[narrow blade]

increasing : (0, pi/2) U (3pi/2, 2pi)
decreasing: (pi/2, 3pi/2)

[chilly pollen]

Good, you found an interval for $\cos x$, now to find one for $\cos 1.5 x$, what do we do?

[soft zealot]

Adhi

[chilly pollen]

i want to you to think this through for a minute

[narrow blade]

since it shrink , it should multiple with x

x can be represent the unit circle

[chilly pollen]

what operation do you do then @narrow blade

[narrow blade]

sorry what does operation means? , is it + - */ 😓

[chilly pollen]

yes

precisely

[narrow blade]

multiply

i got this

i got increasing is right, but for the decreasing something was off

oh, 4pi/3 is < 2pi

got it ❤️

.close

[final saddle]

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[chilly pollen]

. cool stuff

[final saddle]

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[brazen stump]

hello in this question do i assign the ratio as 1x:4x or 1r:4b?

[brazen stump]

and why

[pliant shore]

you should properly define your variables before you use them

so letting x be the number of red counters

4x is the number of blue counters

that approach works

[brazen stump]

what do you mean

[brazen stump]

how about 1r:4b

[pliant shore]

someone reading this will definitely ask: "what is x?"

[pliant shore]

that actually doesn't make sense

[brazen stump]

why

i can do r+4b as the total

[pliant shore]

why are you doing 4 * (number of blue counters)

or what even is b

[brazen stump]

is it bc there is 2 unknown varibales b and r and wheras using x there is only missing on

[brazen stump]

why am i doing 4*x

[pliant shore]

if x is the number of red counters

there are 4 times as many blue counters

so 4x is the number of red counters

[brazen stump]

oh i see

ok thats makes sense

[pliant shore]

yes so you want to get stuff in terms of one variable

[brazen stump]

so you dont do them independtly r:4b

[pliant shore]

yep, that's not correct

[brazen stump]

is there ever a case where u do them independetly

[pliant shore]

no

[brazen stump]

ok thanks south

.close

[final saddle]

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[pliant shore]

well it's more like, if you have r red counters and b blue counters

r : b = 1 : 4 means r/b = 1/4

so b = 4r

[pliant shore]

so I understand where you're coming from

but you should try to get everything in one variable

[brazen stump]

ohh

[pliant shore]

with 2 variables it's more confusing I feel

[brazen stump]

but the other method works aswell?

[brazen stump]

this

[final saddle]

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[pliant shore]

yes, it can work that other way

[pliant shore]

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[keen wolf]

I am having difficulty proving this.

[thin cloud]

You're indian?

Anyway what have you tried

[keen wolf]

I am Nepali

[thin cloud]

Oh cool

[warped pebble]

write cos2A is 1-sin^2A

maybe

[thin cloud]

So, the idea is to use sum to prod for cos(2a) and cos(2b)

[leaden moon]

Oh it's not squared

[warped pebble]

oh shi mb i might be making a mistake then

[thin cloud]

The whole idea is to use sum to prod identity tbh

[warped pebble]

right right

uhhh

[warped pebble]

cos on both sides

cos (A+B+C)

?

would that work?

[thin cloud]

just cos(2a)+cos(2b) then maybe cos(2c)-1

double angle identity

[keen wolf]

cos both sides ??

[warped pebble]

one sec lemme solve and see

[warped pebble]

the question is wrong

it should be minus of 1 + 4sinAsinBsinC

[heady moon]

hint use this

[leaden moon]

,rccw

[heady moon]

eh how'd you do 2

[final saddle]

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.

[keen wolf]

but in rhs we have sin

[thin cloud]

$\sin^2{A}+\sin^2{B}+\sin^2{C}=2\sin{A}\sin{B}\cos{C}$

[soft zealot]

Alexis_Fx

[thin cloud]

your this um

i dunno, different?

[thin cloud]

Just use this...

I have played this game before

[thin cloud]

You haven't used the fact A+B+C=180 degree yet

oh

you did

[keen wolf]

but it would result in cos

not sine

[thin cloud]

hm

Okay wait

,w 1+4sin(30)sin(50)sin(100)

[soft zealot]

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[thin cloud]

,w cos(230)+cos(250)+cos(2*100)

[soft zealot]

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Requested by yobi_fx.

[thin cloud]

lmao

The question isn't right

$\cos(A)+\cos(B)+\cos(C)=1+4\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})$ for $A+B+C=180^\circ$

[soft zealot]

Alexis_Fx

[thin cloud]

Maybe a typo?

Maybe A+B+C=90 instead?

[keen wolf]

don't know but this came in our pervious exam

[heady moon]

check again

share photo if possible

[thin cloud]

Seems like free point to me haha

If the problem is wrong they should give everyone free point for it

[heady moon]

typo for sure

[thin cloud]

Is the first question asking for 🌫️ ^-1(3)

[thin cloud]

Free point!!

[keen wolf]

It's for grade 10 , Guess it would be easy for you guys

[thin cloud]

I mean we have played the game before so kind of

These are pretty tough problems for 10th grade me tho

[keen wolf]

if the sign of 4sinA.sinB.sinC was (-) could it be proven

[thin cloud]

Yeah, just looked that up

Oh wait no

No, it's different

[heady moon]

-cos2C would work

[keen wolf]

if there was -cos2C then rhs would be 1- 4sinA.sinB.cosC

[heady moon]

probably

[keen wolf]

Thanks Guys
And sorry for the trouble

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[keen wolf]

.close

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[steady cobalt]

i need someone to teach me integration by parts i have finals coming up

[left echo]

Do you have a specific instance of it

[steady cobalt]

you mean an example?

i dont but ill ask gpt for some

i just dont understand anything but i know the rule

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[left echo]

I wonder how outdated that prompt is

[heady moon]

lmao fr

[steady cobalt]

ill try to get some examples not from ai then

[hybrid heath]

Your book should have plenty of exercises

[steady cobalt]

we dont have a book

as a matter of fact the doctor didnt explain integration by parts

[hybrid heath]

Brutal class

[steady cobalt]

its just that the head profesor said its in

and our prof didnt both and said deal with it

[soft zealot]

jewels!

[steady cobalt]

x is u and ex is v?

[left echo]

Show me the formula you're working with

[steady cobalt]

oh yes

its

uv - integration v d u

[soft zealot]

jewels!

[left echo]

Right, so you pick u to be a function that makes the second integral doable

In your case, yeah, x = u and e^xdx = dv

[steady cobalt]

because of

liate rule

oh idk what it was i kept watching yt vids

[true matrix]

Well it’s more the liate rule works because it makes it easier

Not it’s easier because of the liate rule

[steady cobalt]

yeah

so now we replace the u and v we have to the equation

which should be

x e power x - integral ex d x

d of x is obviously one

[true matrix]

Yup

[steady cobalt]

but what now

i know the d of e x but not the integral\

[left echo]

Huh?

$xe^x - \int e^x \dd{x}$

[soft zealot]

jewels!

[true matrix]

If you have a function f(x) and then find the derivative f’(x), what happens when you get the integral/antiderivative of f’(x)

[left echo]

You don't know how to evaluate the integral there?

[steady cobalt]

then it gets reversed?

[true matrix]

Basically yeah

So if f(x)=e^x, f’(x)=e^x as you said

So what’ll happen when you integrate it

[steady cobalt]

yeah sorry

[steady cobalt]

then the integral is still\

e x?

[true matrix]

Sure

Plus a constant though

e^x+C

[steady cobalt]

ok so if we get an integral do we just not do anything?

[left echo]

You probably should review basic antiderivatives before proceeding with integration by parts

[steady cobalt]

because we are getting a derivative then just reversing

[true matrix]

No

Well sure if you get a function then differentiate it then find the antiderivative youll be back where you started up to a constant

[steady cobalt]

wait so for example

whats the integral of ln x

the derivative for it is 1 over x

[severe canyon]

Don't think this way

[steady cobalt]

so the integral is ln x?

[steady cobalt]

then how

[true matrix]

The antiderivatuve of 1/x is lnx yes

Again, plus a constant

[steady cobalt]

yes

[left echo]

Most places use an absolute value to match domains

[severe canyon]

Remember that integrating lnx means finding a function whose derivative is lnx

But this is not straightforward to find

[steady cobalt]

wait these profesional words are confusing me

i need to have it applied on a problem

let me get something

[severe canyon]

Which professional words?

[steady cobalt]

$\int \tan^{-1}(x) , dx$

[soft zealot]

onfl

[steady cobalt]

how would someone solve this

tan should be u

and x is v

[pallid thicket]

formula

[true matrix]

Formula 😭

[steady cobalt]

so uv - integral v d u

[severe canyon]

With integration by parts

[true matrix]

Mm like jewels said if you don’t know what the integral of e^x is you probably shouldn’t tackle such problems yet

[severe canyon]

Mmmh

[steady cobalt]

is x not v?

[severe canyon]

Not at all

[steady cobalt]

the doctor said to ignore it because the exam comes with pre instated rules

[pallid thicket]

lol

[steady cobalt]

😭

[severe canyon]

x is not even multiplied

[true matrix]

The doctor is very wrong

Sorry

[steady cobalt]

dont appologize

i realized something is off

[true matrix]

Do you know

$\int x^2 $

Oh well

The integral of x^2

[steady cobalt]

power rule?

x power 3 over 3

[true matrix]

Sure

How about sin(x)

[steady cobalt]

• cos (x)

why is it a dot

[true matrix]

Sin^2(x)?

[steady cobalt]

sin power 2 ?

[true matrix]

Not quite no

[steady cobalt]

no im asking what the question is

[true matrix]

The integral of sin(x) squared

[steady cobalt]

hmm

i dont know

:/

[true matrix]

What techniques have you learned

[steady cobalt]

power rule

[severe canyon]

What is cos(2x)? Do you remember any double angle formula?

(from trig class, not talking about integrals)

[steady cobalt]

-sin(2)?

[severe canyon]

$$\sin^2x \neq \sin(2x)$$
$$\sin^2x \neq \sin x^2$$

$$\sin^2x = {( \sin x)}^2$$

[soft zealot]

Alberto Z.

[severe canyon]

Just to make sure of the basics 😅

[steady cobalt]

yeah thats true

[soft zealot]

Alberto Z.

[severe canyon]

@steady cobalt

[steady cobalt]

double angle formula?

i mean trig class isnt in the finals so i didnt revise it

im starting to think my professor threw us under the bus and im just realizing it

also cant i use

tabulars method

i heard its quite easy

[severe canyon]

Oh ok then maybe your course doesn't cover some integrals

[true matrix]

What sre you studying btw

[steady cobalt]

AI

[true matrix]

Just called AI?

[final saddle]

@steady cobalt Has your question been resolved?

[steady cobalt]

Sorry for late response I had to do something

It's artificial intelligence

It's my first semester

[final saddle]

@steady cobalt Has your question been resolved?

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[crimson cypress]

!status

[final saddle]

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

[crimson cypress]

2

[atomic moon]

What did you tried ?

[tired walrus]

show where you got so far

[crimson cypress]

i've tried following:

1+sqrt(x) = u

sqrt(x) = u

but no luck

[tired walrus]

good start, what did you do after

[atomic moon]

so u - 1

[crimson cypress]

new limits: 2 to 3

[tired walrus]

dx = 2(u-1) du

[atomic moon]

Well since sqrt(x) = u-1 you can replace

[crimson cypress]

yeah

[atomic moon]

Yeah so now its good no?

[tired walrus]

good, this is quite workable now

[atomic moon]

You won

[crimson cypress]

workable?

but

1 over u

[tired walrus]

you can do it from here

[crimson cypress]

sorry -2 over u

[atomic moon]

thats ln(u)

[crimson cypress]

how do integrate that

[tired walrus]

do you know int 1/u du

[crimson cypress]

aaah that's a fact

[tired walrus]

it's famous

[atomic moon]

Take out the 2 and integrate

[crimson cypress]

roger

on it now

is that correct?

<@&286206848099549185>

[static hedge]

whats this?

[crimson cypress]

i tried to plug values on that one

[static hedge]

wtf did u do here?

[static hedge]

This is fine

after this what?

[crimson cypress]

u = 1+ sqrt(x)

[static hedge]

Let me ask you a question

[crimson cypress]

yeah

[static hedge]

What is $ln(a+b)$ in simplified terms?

[soft zealot]

Kavy

[crimson cypress]

lna * lnb

[static hedge]

WOW!

Ok then; What is $ln(a) + ln(b)$?

[soft zealot]

Kavy

[static hedge]

Get yr basics right!

[crimson cypress]

you're right

i forgot these simple things

[static hedge]

🙂

[crimson cypress]

lna + lnb = ln(a*b)

[static hedge]

Rest is correct

[crimson cypress]

what's the answer to that tho?

[crimson cypress]

that is correct then?

[vital crag]

,w int 1 to 4 1/(1+sqrt(x))

[soft zealot]

Results provided by WolframAlpha

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[crimson cypress]

alright

thanks!

.close

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[robust sequoia]

where did I make a mistake

[robust sequoia]

why is the image flipped 😭😭

[scarlet sequoia]

This is already sus

The second term has a wrong sign

-e^.../... - (- 1/....)

[robust sequoia]

wait can you circle it please

[scarlet sequoia]

There are 4 terms in total

$-\frac{1}{2\pi n i}$

[robust sequoia]

oh i see it now it should be + not minus

[soft zealot]

Raphaelisius Maximus MMIII

[robust sequoia]

thankssssss it was such a dumb mistake

.close

[final saddle]

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[robust sequoia]

back cuz I can't solve question 2 🥀

.reopen

[final saddle]

✅ Original question: #help-36 message

[robust sequoia]

how do I isolate step(t) from the integral

[left echo]

It seems like they did it

[robust sequoia]

no like am i supposed to find that expression of step(t) or is it given to find the other expression ?

they say show that step(t) =..... so i need to find it first

should i use the definition of the integral ?

[steady locust]

what video is this from

for context

[robust sequoia]

3b1b about fourier series

https://www.youtube.com/watch?v=r6sGWTCMz2k&list=PLZHQObOWTQDNPOjrT6KVlfJuKtYTftqH6&index=4

[steady locust]

the step function used here is 1 for (0,0.5) and -1 for (0.5,1)

you can see in the video

the step function is a family of functions, this is the specific one here

you can also see that in how the integral is decomposed

[robust sequoia]

yes but how can this help in finding the expression

[steady locust]

wdym?

[robust sequoia]

step(t)=.....

sorry if i sound dumb 😭

[steady locust]

you found c_n right

[robust sequoia]

yes

[steady locust]

the first equality in the line step(t) = ... is given, if that was not clear

plug in c_n, see how you can tranform that sum

[robust sequoia]

ohhhhh i thought i had to find it

this makes things so much clearer

thanks !

.close

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[lavish gyro]

for this frq, im confused on two parts

firstly:

on part b, im confused on how u find out that the solution is concave up? how do they know its above the like y=1/2x-1/2

second:

on part d, after m=(m+1/2)x+b-1

how do they know

0=m+1/2 and m=b-1

[next ether]

For the first part they said that if y≥-1/2 x+1/2 , y will be concave up(because in that case d²y/dx²≥0) otherwise it's the opposite

[lavish gyro]

so ur saying above the line, d^2y/dx^2 is postiive,

but how do they know y>1/2x+1/2? am i missing a part, bc how do they get from the second derivative to that?

[next ether]

For the second part if you have an equation like ax+b=cx+d for all x in R then you have a=c and b=d

[lavish gyro]

what do u mean by in R?

and for ax+b=cx+d, is that an algebra rule or smht, bc how do u know that they are equal?

[next ether]

The second derivative of y is y+1/2x-1/2 so in order to have this quantity to be positive we need y>-1/2x+1/2

[lavish gyro]

why must only the y value need to be postiive for it to be CCU, why not the x value?

[next ether]

R is the set of real numbers
And basically for the second one the thing i'm trying to say is that if two affine functions are equal then their slope and the y-intercept are equal

[next ether]

The y value don't need to be positive we just want the x values such that y>-1/2x+1/2

[lavish gyro]

oh okay, i see, so basically the "like terms" are equal

how do u know its equal to 0

for 0=(m+1/2)x

[lavish gyro]

okay i see, so we are making sure that y is greater for any value of x u plug into the equation, ensuring its CCU?

[next ether]

In this case we have m=(m+1/2)x+b-1 we can rewrite it as 0x+m=(m+1/2)x+b-1
So 0=m+1/2 and m=b-1
Therfore m=-1/2 and b=m+1=1/2

[next ether]

Yes

[lavish gyro]

oh i see, its equal to 0 because there is no "x" on the other side

right?

[next ether]

Yes

[lavish gyro]

okay thank you!

.close

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[civic furnace]

A bicycle wheel has mass M and radius R and can rotate about the axle
through its center. A spring of spring constant k is connected to a spoke
on the wheel a distance r < R between the axle and the rim. The spring is
held horizontal and the other edge is connected to a wall. When the contact
point between the wheel and the spring is directly above the pivot point,
the spring is unstretched. The wheel is then turned slightly and released,
initiating small, simple harmonic oscillations.
a) Find an expression for the angular frequency of the wheel using energy
or force/torque ideas.
b) Find an expression for the period of oscillation for the wheel.
c) Repeat a) and b) using whichever one you didn’t use in a)
d) Explain conceptually how the oscillation would change if the spring were
connected (still horizontally) at the edge of the wheel instead of at r.

[civic furnace]

would this be correct?

[loud sundial]

discord.gg/physics

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[untold kiln]

I would like an opinion about a passage from a book. The image is a stitched version of the longer passage, to include only the relevant parts.
I included the parts for context. What I'm curious about is the part marked with a pink star; specifically, the description of the underlined part.

[untold kiln]

To me, the math clearly says that the value of (P-E) . N is equal to the vector from E to P, projected onto N. However, the description shown doesn't read to me like that.

First of, I find the description of that mass convoluted. A big runon sentence.

What I read the description as saying is tha the length of that (P-E).N value is from P to Pnew, projected onto N.

Does anyone else read the description like I am?

Specifically, this part: Notice that since the mass (P − E) ⋅ N of the mass-point P new is the signed distance from the point P to the plane through the eye point E parallel to the plane of projection S,

Basically, I think that description is both convoluted and wrong.

Or, is it me that's not seeing the genius of the author?

btw, I think that (P-E).N ends up being the line segment marked blue, projected onto N to endup as the yellow line.

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@untold kiln Has your question been resolved?

[opal pelican]

Since, due to the mention of mass and the notation, this seems like a physics problem, you'll probably have more luck asking in the physics server

[opal pelican]

.

[untold kiln]

This is in the context of quaternions. It's computing perspective projection with quaternions.

But, I'm asking for a reader to judge on what they see on the page; no external info necessary.

should be self contained.

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@untold kiln Has your question been resolved?

[untold kiln]

.close

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[livid tangle]

I need someone to explain how to solve this question please

[wet egret]

Ok well do you know what surface area means

[livid tangle]

Yeah I know the formula

[wet egret]

Ima ask u painfully obvious questions till u get it

[livid tangle]

Its 6×edge length right?

Oh sorry mb

[wet egret]

Thats for a cube, yes

[livid tangle]

6×(edge lenghth)squared?

[wet egret]

Ima take that as a yes

[livid tangle]

Oh yeah sorry I know what surface area is

[wet egret]

Lowkey ima let the other nerds take over

They know more about explaining then i do

[livid tangle]

Okay 👍

[fossil kiln]

nerdsTT

[viral jungle]

consider calculating the surface area of the solid separately from all 6 sides

specifically, if we call the current angle the front, the front and the back will have the same surface area

[livid tangle]

So like 6×(3)^2?

[viral jungle]

why 6?

[livid tangle]

Cuz that's part of the formula for finding the surface area no?

[rocky tusk]

instead of blinding using formulas you should understand what you’re actually calculating

[viral jungle]

that only applies if you have an entire cube exposed

[fossil kiln]

what would be the surface area of one face ie a square

[rocky tusk]

6s^2 comes from the fact that there are 6 faces of a cube each of which having area s^2

[livid tangle]

Okay

[rocky tusk]

you want the surface area of the surface which means you want to count the number of exposed faces then multiply by s^2 (the area of each face)

[livid tangle]

Ohh okay I think I get it now

[rocky tusk]

there are many ways to count here

koyuki offered one way

[livid tangle]

Should I just count the total faces?

[rocky tusk]

you could do a brute force way if you insist

[rocky tusk]

exposed faces

[livid tangle]

Yeah the total exposed faces mb

The front has 9 faces so the back should have 9 too

So if I find the surface area of the front that would be 9×(3)^2?

[viral jungle]

yep, sure is

and you've also realized that the back has the same number

[livid tangle]

Yep

[viral jungle]

keep going

[livid tangle]

So the surface area of the front is 81

And the back is also 81

[viral jungle]

in fact, I think you are now equipped to do the rest

you just need to find how many faces are exposed from each direction

[livid tangle]

But what about the hidden faces should I find those too?

[viral jungle]

warning: this is NOT the same as counting the number of faces you can see if you look at the solid from one side

[rocky tusk]

i mean you can count the hidden faces then subtract it from 6 times the number of cubes (total number of faces)

[wet egret]

No

[livid tangle]

Okay so If I just find the exposed faces I dont have to find the hidden faces then?

[rocky tusk]

sure

[viral jungle]

again, recall what you're being asked to count

[rocky tusk]

but this is actually the way i would’ve done it tbh

[viral jungle]

you're being asked to calculate the surface area of the solid

whichever way you use, the hidden faces are of no use to you because they are NOT part of the surface of the solid

[livid tangle]

I was initially trying to do it that way too but I got confused

So from the top there are 5 faces

So the surface area should be 210

And the bottom will also be 210

And from the left there are three faces exposed so it should be 54

And from the right there are 4 faces exposed so it should be 84?

[livid tangle]

Is that not it?

[livid tangle]

Ohh right

[viral jungle]

I warned you for a reason

[livid tangle]

Sorry

So from the left its 84 again

Okay now that I have all these what do I do now?

Do I just add them all up?

[viral jungle]

well what do you think?

[livid tangle]

I think I should add them all up

Cuz they're the different sides of the figure and I need to find the total surface area right?

[viral jungle]

the word total is a big hint

[livid tangle]

So I don't add them?

[viral jungle]

you do

[livid tangle]

Okay

[viral jungle]

by now words like "sum", "total", etc. should be a big hint

[livid tangle]

It came out 540

Is that right?

[viral jungle]

I didn't bash the numbers, but assuming you didn't make some oopsie somewhere, it should be

[livid tangle]

Thank you soo much I was stuck on this question for a while

I was trying to do the hidden faces method but it was just confusing

[viral jungle]

the hidden faces method does require you to keep track of each cube's hidden faces separately

might be confusing to some, but not hard once you get the hang of it

[livid tangle]

Okay but if I were to do it that way

There are 9 hidden faces right?

And one faces has two sides?

[viral jungle]

if you remember to multiply the number of hidden faces by 2 then yes

[livid tangle]

Wait

There are 9 hidden faces

And If I multiply them by 2 that's 18

And the total surface area is 486

[viral jungle]

let me see again

[livid tangle]

Sorry but Im getting confused again

You want me to send the question again?

[viral jungle]

wait I only counted 8 faces

where did you get 9

8 hidden*

[livid tangle]

How