help-43

[serene junco]

from there, you can turn one positive solutions into 3 negative solutions

[deft tangle]

Ohh i got it

My bad

[serene junco]

like this

[deft tangle]

Thank you very much

.close

[compact pewter]

Channel closed

Closed by @deft tangle

Use .reopen if this was a mistake.

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[keen glacier]

if I have

d = a, and 1 + d^2 = b, what's the relation between a and b

[torn stratus]

just put the values and find it?

[serene junco]

since d = a,
1 + a^2 = b

and 1 + a^2 = b is the relation between a and b

[keen glacier]

by the way, assume a and b are given

[serene junco]

if you want, you can rearrange it to
b = 1 + a^2
or
a = sqrt(b-1)

[torn stratus]

!noans

[compact pewter]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[civic quartz]

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[serene junco]

is this directed at me

[torn stratus]

yes

[serene junco]

, well im not the one asking the question

[torn stratus]

"not to hand out answers."

[soft bobcat]

That command is !nosols

[torn stratus]

thnku

[serene junco]

the answer was one step away for this question so i don't know what else i could have done

[torn stratus]

i mean ya 💀

[compact pewter]

@keen glacier Has your question been resolved?

[keen glacier]

wait a bit

im giving the entire question

so the way it looks right now is that the relation b = 1 + a^2 does not hold. b is 17/4, and a is 1, right? so what I wanna scale the expression and get all possible values of d where the relation holds (answer is d = 4, and d = 1/4)

how should I scale the expression?

[serene junco]

what are a and b here

[keen glacier]

b is 17/4, and a is 1

[serene junco]

i don't see how changing c and d would affect that

[keen glacier]

I wanna scale the expression up and down so that b = 1 + a^2 holds true

what equation do I use

[quartz yoke]

Can someone tutor me trigonometry ln Monday

AHHHH

Why is everyone so busy

[devout latch]

#❓how-to-get-help

[quartz yoke]

Thanks

[compact pewter]

@keen glacier Has your question been resolved?

[keen glacier]

so according to the expression above, d^2 + 1 = 17/4, but at the same time, d = 1, which doesn't make any sense

[gritty shuttle]

nah

not 17/4=d^2+1, but 17/4=A(d^2+1)

[keen glacier]

I don't wanna use A. I wanna know to scale this to get the correct relation. The answer is scale it by 4 and 1/4

but how did they get 4 and 1/4

is there an easy way to know

[gritty shuttle]

maybe using A is easiest

solve a system of 17/4=A(d^2+1) and 1=Ad

[keen glacier]

wow

how did u get that

[gritty shuttle]

u probably comparing coeffs and consists right?

[keen glacier]

but wait a second

yes

i wanna scale the expression by some factor so that the relation b = 1 + a^2 holds true, b in the case above ^^ is 17/4, a is 1

is there some equation I can use to get the scaling factor

[gritty shuttle]

a (17/4) is a constant of lower left expression and b is a coefficient of lower left exp
correct?

[keen glacier]

yes

[gritty shuttle]

ahh

[keen glacier]

nope actually

the opposite

[gritty shuttle]

what

[keen glacier]

b is 17/4

a is 1

b is the constant

[gritty shuttle]

ah

[keen glacier]

a is the coefficient

wait let me write it again

[gritty shuttle]

i feel like I'm still misunderstanding something...

mmm

[keen glacier]

b is 17/4 and a is 1,

b is the constant

[gritty shuttle]

ahh

[keen glacier]

it's 4 or 1/4, but how can I get this from scratch

is there some equation that can give me this?

[gritty shuttle]

this

umm

what are you stuck on

[keen glacier]

if u multiply it by 4 top and bottom for example, it would give u a = 4, b = 17, b = 1+4^2 = 17

how did they get 4? that's where im confused

i definitely can't just guess it

[gritty shuttle]

$\frac{X}{Y}=\frac{x}{y} \iff X=AY and Y=Ay$

[boreal girder]

へいほー

[gritty shuttle]

you know

[keen glacier]

what's this

[gritty shuttle]

Expressions about fractions

by using Y=Ay, in this case, we get 17/4-(z+z^(-1)) = A(1+d^2-d(z+z^(-1))).

and

[keen glacier]

I want b = 1 + a^2 to hold, my a is the coefficient and my b is the constant

how can we write a simple equation

that checks for when b = 1 + a^2 hold

and what factor to multiply C(z) by

[gritty shuttle]

why do you stick to b=1+a^2?

bcause 1+d^2?

[keen glacier]

yh

[gritty shuttle]

wait

thinking about an explanation

mmm

[keen glacier]

what's the equation u derived?

[gritty shuttle]

as this equivalence transformation implies, when considering the equality of fractions, you have to consider the ratio A.

[keen glacier]

what's X, Y, x, y, and A

[gritty shuttle]

any

like 1/3 = 5×1 / 5×3 = π×1 / π×3

ah but not 0

[keen glacier]

Ok so here what do we have

[gritty shuttle]

so in this case, 17/4-(z+z^(-1)) not always eq to 1+d^2-d(z+z^(-1))

[keen glacier]

Yes. Here, it's not equal. I want it to be equal.

[gritty shuttle]

since we want the fractional equation to be an identity, we use the Y=Ay (or BY=y) relationship we just discussed, and make Y=Ay an identity accordingly.

[keen glacier]

what's A, B, and y here

in my example

[gritty shuttle]

Y = 17/4-(z+z^(-1) , y = 1+d^2-d(z+z^(-1)), A is some const.

[keen glacier]

can u write it in latex

[gritty shuttle]

ok

$Y = \frac{17}{4}-(z+z^{-1}) \ y = 1+d^2-d(z+z^{-1}) \ A \in \mathbb{R}$

[boreal girder]

へいほー

[keen glacier]

What's X?

right?

wait it's wrong

[gritty shuttle]

just comparing these

upper left is X, lower left is Y, upper right is x, and lower right is y

[keen glacier]

ok what are X and x

[gritty shuttle]

you

$X=5-2(z+z^{-1})\x=K\left(1+c^2-c(z+z^{-1})\right)$

[boreal girder]

へいほー

[gritty shuttle]

for this question, there is no need to think too much about X and x.

[keen glacier]

ok, now how do we get what we need to get

let me see

[gritty shuttle]

as i said, Y=Ay.

wait i use BY=y instead sry. B is also some const

so

$B\left(\frac{17}{4}-(z+z^{-1})\right)=1+d^2-d(z+z^{-1})$

[boreal girder]

へいほー

[gritty shuttle]

by comparing const (B×17/4 = 1+d^2) and coeff (B = d)

we get

$\frac{17}{4} B = 1+B^2\ \iff (4B-1)(B-4)=0\ \iff B=4 \quad or \quad 1/4$

[keen glacier]

nice

[boreal girder]

へいほー

[keen glacier]

and d is just B right?

[gritty shuttle]

yeah

[keen glacier]

how did u figure this out

[gritty shuttle]

just equivalence transformation

equivalence transformation is the essence of math

[gritty shuttle]

like this

[keen glacier]

so in my original C(z), the denominator did not equal 1 - dz - dz^-1 + d^2, right

[gritty shuttle]

uh

yea

[keen glacier]

Ok cool

thanks for ur help

.close

[compact pewter]

Channel closed

Closed by @keen glacier

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[quartz yoke]

Integrate 1/(x-a)(b-x) where b>x

[quartz yoke]

!status

[compact pewter]

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

[mortal nimbus]

Uhh

Just

[quartz yoke]

I got a diff ans from whats given

[mortal nimbus]

Write the numerator

As

[quartz yoke]

I used cts

[mortal nimbus]

(x-a)+(b-x)

[quartz yoke]

And applied dx/root of 2-x2

[ruby mason]

partial fractions are probably your friend here

[mortal nimbus]

No no

[mortal nimbus]

Then manipulate

By dividing with

b-a

Then its done

[ruby mason]

this isn’t even equal to 1

wtf

oh

[mortal nimbus]

just divide

With b-a

[quartz yoke]

Trev

U doin jee prep right?

[mortal nimbus]

Yea

[quartz yoke]

Wait wait

I wrote the question incorrectly

[mortal nimbus]

[quartz yoke]

Its 1/whole root of (x-a)(b-x)

Sorry 💀😭

[mortal nimbus]

Substitution

[quartz yoke]

I used cts

[mortal nimbus]

Easy substitution

[quartz yoke]

Umm

[unkempt sail]

I think the sub was something like acos^2 theta + bsin^2 theta

[mortal nimbus]

U could

Just

Yk

Nvm

Let him learn

[unkempt sail]

This typa sub is very weird

[mortal nimbus]

There is an easier

[unkempt sail]

I doubt he would be able to guess it

[unkempt sail]

Could be

That's my idea though

[mortal nimbus]

Its just a 1/root quad

[quartz yoke]

Idk what sub yall talkin abt

I used cts and used sin inverse of x/a

[unkempt sail]

Wha

[mortal nimbus]

Hmmm

[unkempt sail]

What is cts first off

[mortal nimbus]

Completing the square

[ruby mason]

yeah that checks out

[unkempt sail]

Hey that works

Yeah

[ruby mason]

why would you ask if you knew the soln tho

[quartz yoke]

No i got diff ans

[unkempt sail]

!show

[compact pewter]

Show your work, and if possible, explain where you are stuck.

[quartz yoke]

Showing work is not possible

But i can tell u my ans

Sin inverse of (2x-a-b/a-b) is what i got

Ans is 2 sin inverse of root of x-a/b-a

Basically how do i change my ans to given ans

[mortal nimbus]

Think u did some calc mistake somewhere

[quartz yoke]

So heres work

Is smth wrong?

[mortal nimbus]

Something is wrong here

Im pretty sure

Too lazy to check

[quartz yoke]

Bruh

[mortal nimbus]

I have done the same integral before

Like 2 months back

[quartz yoke]

Are u 12th or?

[mortal nimbus]

12th

[quartz yoke]

We both are the same age

Yet u r a helper and im a seeker

Lol

[mortal nimbus]

I also asked a question

[molten badger]

Helper can be at any grade xd

[quartz yoke]

So whats wrong with the work?

[mortal nimbus]

Im too last to do

Check again

Properly cts

[quartz yoke]

Ye i meant hes on a another lvl when compared to me

[mortal nimbus]

Keep up the prac

[quartz yoke]

U do jee

Well i dont

[quartz yoke]

I checked thrice and i dont see anything

Checked*

[mortal nimbus]

Then i might be wrong

And the key is wrong

[quartz yoke]

Its rd sharma tho

Well i spotted minor errors but not the whole ans

Lemme check rq

Umm

They used u sub

Subbed x-a as u^2

So by cts is my ans right?

[molten badger]

What's the key btw

I think so?. I'm not so sure

[quartz yoke]

,rccw

[quartz yoke]

5th

.close

[compact pewter]

Channel closed

Closed by @woven loom

Use .reopen if this was a mistake.

[quartz yoke]

.reopen

[compact pewter]

✅

[quartz yoke]

Integrate 1/whole root of 5x^2 - 2x

Uhh

[mortal nimbus]

Move on from ncert

😔

[mortal nimbus]

Just complete the square

[quartz yoke]

I aint preparing for jee

[mortal nimbus]

Oh

Then stick to ncert

😋

[quartz yoke]

This is rd sharma not ncert

[mortal nimbus]

Rd sharma is ncert level😔

[quartz yoke]

Bruh

For u it is yes

[mortal nimbus]

i mean practice 600 integrals in a week

[quartz yoke]

For what?

Integration isnt the only chapter i have for class 12 exam

[mortal nimbus]

Oh

Ur not giving jee

😔

[quartz yoke]

Les quit talking and how abt u help me

[mortal nimbus]

Cts

For the question

[quartz yoke]

Ans is kinda slightly different from mine

Anything wrong here?

[mortal nimbus]

Uhhh

[quartz yoke]

K forget

.close

[compact pewter]

Channel closed

Closed by @woven loom

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Available help channel!

Send your question here to claim the channel.

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

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

[young tinsel]

hi

[young tinsel]

Problem: Simplify $\frac{\dbinom{200}{75}}{\dbinom{200}{74}}.$

[boreal girder]

7

[young tinsel]

I tried rewriting the expression using pascal's identity

so (199 choose 75 + 199 choose 74)/(199 choose 74 + 199 choose 73)

and then made it equal to x

and then i multiplied and distributed

so 199 choose 75 + 199 choose 74 = (199 choose 74)x + (199 choose 73)x

i dont know what to do next pls help

however, i notice that the denominator 200 choose 74 is equal to 199 choose 75 - 199 choose 73

<@&286206848099549185>

[devout latch]

if you want to evaluate this, just use the definition of the binomial coefficient?

[young tinsel]

how

[devout latch]

write it out. you will get a fraction of fractions with a bunch of factorials canceling out

[young tinsel]

ok

thnak you

for helping

.close

[compact pewter]

Channel closed

Closed by @young tinsel

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Send your question here to claim the channel.

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

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

[civic glade]

Can sb give me some ideas, thank you

[signal willow]

<@&268886789983436800>

this looks like a scam ngl

[keen granite]

nah wdym, totally legit!!

[signal willow]

Maybe consider the determinant as a function of the diagonal entries. It's a pretty nice form.

Then, you can by a proof by contradiction, assuming all ways give a singular matrix and deduce that the coefficients of this function are all 0. However, also calculate one of the coefficients and deduce a contradiction.

[quartz yoke]

Guys The first to solve this question will win!

Diffrentiate e^x using chain method

[signal willow]

!occupied

[compact pewter]

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).

[river portal]

I think you can do induction. First consider replacing all diagonal entries with 0. Think about what is implied about the products in the determinant of the original matrix if the resultant matrix is still not invertible. Then try having the first diagonal entry as 1 and others as 0

[signal willow]

quite similar ig

[compact pewter]

@civic glade Has your question been resolved?

[civic glade]

Thank u so much, I'll try again

[compact pewter]

@civic glade Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[river portal]

.reopen

[compact pewter]

✅

[river portal]

Any progress?

[compact pewter]

@civic glade Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

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

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

[visual juniper]

A nine digit number is formed by writing all the numbers from 1 - 9 in random order. What is the probability that the number formed is exactly divisible by 4

[kind viper]

!status

[compact pewter]

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

[visual juniper]

i got an answer

actually

i got an answer

i asked chat gpt

[glass turret]

!nogpt

[compact pewter]

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).

[visual juniper]

chat gpt gave me a different answer

[kind viper]

!nogpt

[compact pewter]

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).

[visual juniper]

Im aware of that

u told me that last time

[glass turret]

chatgpt is pretty bad at questions like these

[visual juniper]

Im just saying for reference

[kind viper]

dont trust gpt for shit, dont even look at its answer

[glass turret]

i wouldnt trust it

[kind viper]

the fact it didnt match yours doesnt mean SHIT

[visual juniper]

I tried to solve it again

this time

[kind viper]

show your answer and your work

[visual juniper]

I got a DIFFERENT ANSWER

show my work uh...

alright

[glass turret]

actually if you just ask it the same thing again youll probably get a different answer from chatgpt

[visual juniper]

ik

[visual juniper]

theres this one JEE question that I tried to ask chatgpt atleast 10 times but got different (wrong) answer every single time until I used deep research

[glass turret]

yea i wouldnt trust any AI at all

for solving math problems

[visual juniper]

I realized

16 / 9! is wrong for various reasons

the answer that I currently have is 16/81

so

to help u understand.

[kind viper]

your numerator is correct but your denom is not

[visual juniper]

I realized that the last number must be even for it to be divisible by 4, so there are 4 possible even numbers

[kind viper]

there are 16 pairs of digits you could have at the end of your number to make it divisible by 4.

you counted them

[odd]2, [even]4, [odd]6 and [even]8

[visual juniper]

For every possible even number I have listed the number of possible preceding even numbers in brackets

I hope that makes sense

[kind viper]

the total number of two-digit endings is not 9! nor 81 but 9 * 8

[visual juniper]

for which it is divisble

[kind viper]

which is what you should have in the denominator

[visual juniper]

bruh what

16/72?

uh

8/36

4/18

2/9

[hot epoch]

yo

[visual juniper]

2/9

wait that was the first answer i got...

how'd i get the first time?

ohhh yeahh

i remember

but i think thats wrong, lemme check

[visual juniper]

2/9 is the answer I initially got when I only took preceding numbers to be even but I believe that is wrong because the preceding numbers don't necesarily have to be even

[kind viper]

dont confuse "number" with "digit"

[visual juniper]

ok by preceding number what I mean is the digit in 10's place

I think its 16/81

The factor that determines where the number is divisible by 4 or not is the last two digits Correct?

[kind viper]

should i repeat myself

[visual juniper]

yes pls

[kind viper]

yes correct. your denom is still wrong.

the total number of two-digit endings is not 9! nor 81 but 9 * 8

which is what you should have in the denominator

[visual juniper]

u said 9*8

but how did u arrive there

[kind viper]

yes i said 9*8

[quartz yoke]

cool

[visual juniper]

OH

[kind viper]

9 choices for the units digit

[visual juniper]

two digit endings

[kind viper]

then 8 for the second

yes

[visual juniper]

Ohh

right

OHHH

WAIT

I GET IT NOW

AHH I TOTALLY forgot that

the chance of the fi rst digit being what you want is 1/9 but the chance 2nd digit being what you want is 1/8 since you have one less number to worry about now

ah, thanks

Im in 11th and I have a maths competetion to attend next month with 12th's portions so im cooked

anyway thx for ur help

[molten badger]

Heyyy, I'm also 11th grade and will have a math competition test XD

[visual juniper]

which country u from

[molten badger]

Not next mouthn tho

[molten badger]

I'm Vietnamese xd

[visual juniper]

cool

im indian

i suck at permutation combination and matrices

those topics haven't been taught to us but i've tried to learn it on my own and i failed

I've been able to learn integration, complex numbers and stuff

[molten badger]

Dude, no offense but there are so many Indian in this server

[visual juniper]

lol

why would I be offended at that

tbh you could tell me india is trash country and i still wouldn't be offended unless u racist or smth

[molten badger]

Indian are really good at math i guess

[visual juniper]

not rlly

its a misconception

[molten badger]

My competition focus on solid geometry and differential

[visual juniper]

half the people in my class lack skills in maths

like we are learning how to add and subtract complex numbers and the dudes beside me are failing basic addition subtraction 🌚

🌚

[molten badger]

In fact I'm not taught Matrices and complex numbers in high school

[unkempt sail]

He’s not talking about the general public

[visual juniper]

matrices will be taught next year

[unkempt sail]

He’s talking abt peeps doing olys and stuff

[visual juniper]

right, in that case, yes, indians are very good at maths

tbh every country has that though

i think the reason we got good people at that is cuz of our absurd population

[unkempt sail]

Not when you compare the populace of people like them

[verbal geyser]

true

[molten badger]

Oh yeah, that makes sense

[visual juniper]

alright

cya

mhm

nvm

.close

[compact pewter]

Channel closed

Closed by @visual juniper

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[rose goblet]

Anyone knows why this is correct only when b=1 but for all n multiple of 2 and all a

[rose goblet]

Problem is if b is 0 the whole thing is 0 but it shouldn't be

Idk if i have forgotten something st the roots at the start

Cause i cant see a step thats not correct thats why im surprised how i got wrong result

[kind crane]

this is only useful to your partial fraction decomposition when b is not zero

[kind crane]

when b = 0, I is trivial

[rose goblet]

Ok

But

Even when b isnt 0 for some reason there is a mistake

In these steps

Idk where though

Assume b=/0

[kind crane]

try writing the solutions for n=3 and n=4 and see where your assumptions break down

[rose goblet]

It doesnt

Oh wait wdym

This solution is for n multiples of 2

Thats how i defined the problem

[kind crane]

what does "b = 1 but for all n multiple of 2" mean

[rose goblet]

For n odd i cant write sum of n/2 +1 terms

[rose goblet]

Like if i use any multiple of 2 for n

Or any number for a

Its correct

This was the goal

But i also wanted it to be true for any b

Be its only true for b=1

[kind crane]

what does "any multiple of 2 for n" mean

[rose goblet]

Like {2,4,6,8....

[kind crane]

those are usually called even numbers

you can say even positive numbers

[rose goblet]

Yes sorry forgot name in English

For any positive even number

n

And for any real number a

But b=1 otherwise it fails

But it shouldn't since steps seem correct idk why it does

I forgot something somewhere and cant find now

[kind crane]

yes you assumed n was even here

for n=3, you should write out the 3rd roots of unity to see which term you're missing

[rose goblet]

Yes ik

That's not an issue

Thats assumtpion at the start of the problem

So n is even positive, a is real and b is positive

The issue is that the end result only holds for b=1 and no other positive number

[kind crane]

[rose goblet]

Idk why that is

[rose goblet]

Idk why you focusing on n leav n

b is the one that needs fixing

[rose goblet]

Somewhere there needs to be an extra or less or many b terms in that expression

[kind crane]

it's in your first message

[rose goblet]

Bro pls dont argue

You understand what im asking now

[kind crane]

just answering your quesiton

[rose goblet]

Just focus on it now that you understand what i wanna find

[kind crane]

lmao

[rose goblet]

Since assumption at beginning was n even positive a real and b posutive end result should be correct for those assumptions

But its only for b=1

So there is a mistake or many in one or more steps

Thats what im looking for

[kind crane]

no really it's not clear

you haven't demonstrated why your expression is wrong for b not equal to 1

[rose goblet]

Because desmos says so

I would assume its accurate

So by comparison its wrong

If end result is wrong there is a step in the solution that has a mistake

[kind crane]

how do you know you plotted it correctly in desmos

both the correct answer and yours

[rose goblet]

I put my result

And d/dx in the front

Then a/(x^n +b)

When moving a,n no problem

When moving b problem

Ive checked it many times id assume i didnt miss type something all those times and got the same wrong result

Anyone?

[compact pewter]

@rose goblet Has your question been resolved?

[compact pewter]

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[quartz yoke]

,,\int (x-3),\sqrt{x^2+3x-18},dx

[boreal girder]

!

[bleak dock]

!show

[compact pewter]

Show your work, and if possible, explain where you are stuck.

[quartz yoke]

,rccw

[quartz yoke]

Is this right?

[fossil summit]

Hint: U sub

[quartz yoke]

What

Sub which ?

[fossil summit]

WAIT no, I thought I saw a fraction there, mb

[quartz yoke]

Oh k

[eternal pulsar]

First term looks good

I presume you used arcsinh for the rest of it

Just check with WA

[quartz yoke]

Uuh no

[eternal pulsar]

,w integrate (x-3)sqrt(x^2+3x-18)

[boreal girder]

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[fossil summit]

I think there needs to be 729/16

[eternal pulsar]

So that's fucked

[fossil summit]

Ah yes

At the log

Brutal

[eternal pulsar]

They have it

[quartz yoke]

Bruh what method is that

[eternal pulsar]

Just didn't distribute everything

[boreal girder]

;(

[fossil summit]

I meant for the attempt in the photo

[eternal pulsar]

Yeah

[eternal pulsar]

^

[quartz yoke]

,rccw

[quartz yoke]

11

[eternal pulsar]

My eyes have been permanentely bleached

I cannot with these equivalent forms

I know what WA did and I hate it

Lemme crosscheck

,w is 1/3(x^2+3x-18)-9/2((2x+3)/4)=1/24(8x^2-30x-225)

[boreal girder]

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[eternal pulsar]

@quartz yoke You're correct

Jesus I hate wolfram

[quartz yoke]

Ooh

But my book doesnt match tho

[eternal pulsar]

Yeah it might be wrong

[quartz yoke]

Oh

Ok thanks

.close

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[cyan lava]

yo quick question can this be done

[past ether]

Anti derivative of Ln (5x)

do you know it?

[cyan lava]

nop

[kind crane]

log is undefined for negative x, so depends what tools you've been taught

[past ether]

oh yea it was negative..

[cyan lava]

this the full question

[storm mulch]

Yeah but ln(-5) doesn’t exists

[odd lake]

i dont know what course this is so the methods they may use might be different but the natural log function is undefined for values less than or equal to zero, so the question is nonsensical unless they actually want you to go to the complex world

[cyan lava]

yh im just goin to leave it i have no clue like ive been asking AI and it tells me that the integral is undefined

[kind crane]

Yes for this reason

,calc log(-0.5)

[boreal girder]

Result:

-0.69314718055995 + 3.1415926535898i

[storm mulch]

Complex world 😱

[stuck haven]

woa

[kind crane]

#❓how-to-get-help

[stuck haven]

so basically

[past ether]

alr

[stuck haven]

the domain doesnt support

it does NOT stay defined on [-1,0]

so yeah thats a bitchass integral

[cyan lava]

aighttt

thx guys

.close

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[coral iris]

why is it for unit circle we alwayrs write (sinx,cosy)

[coral iris]

like for the points

i mean (cosx,siny)

[past ether]

using pythagorean theorem on the unit circle

[coral iris]

huh

[kind viper]

bad notation.

you mean x = cos(theta) and y = sin(theta) yes?

[coral iris]

yes

the same thing really

[kind viper]

not the same thing

[coral iris]

(costheta,sintheta)

[kind viper]

(cos(x), sin(y)) sounds like a way different thing

anyway!

[past ether]

x is a variable, theta is mostly for angles

so yea

[kind viper]

this is convention.

the idea behind the unit circle is that the zero point for your angle is (1, 0).

[coral iris]

huh?

[kind viper]

and theta increases counterclockwise.

[coral iris]

yea

[kind viper]

i don't think there is really a sacred meaning to why this starting point and direction was chosen.

rather than e.g. (0, 1) and clockwise.

[coral iris]

how can you use it in radinas without having to convert

all the time

[kind viper]

wdym "use it in radians"

[coral iris]

like if you ahve it in radians and need to find a point

like between for a b or c

and do it instead of having to convert

[kind viper]

i have no idea what you're talking about.

are you looking at a problem right now?

cause i really legitimately cannot understand what "have it in radians" or "between for a, b or c" means!

[coral iris]

like this

[kind viper]

ok right so you want to be able to tell which quadrant an angle lies in when it's given in radians.

[coral iris]

yes

[kind viper]

a full circle is 2pi, so a quarter of it is pi/2.

[coral iris]

and not have to convert

[kind viper]

the breaking points between the quadrants are 0, pi/2, pi, 3pi/2, 2pi, ...

(which are just 0*pi/2, 1*pi/2, 2*pi/2 and 3*pi/2)

compare your angle (11pi/6 in this case) against each one of these in turn

[coral iris]

huh?

i mean 11pi/6 is greater than 3pi/2 looks like it

so it is c

.close

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[coral iris]

.reopen

[compact pewter]

✅

[kind viper]

yes, 11pi/6 > 3pi/2

[coral iris]

.close

[compact pewter]

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[kind viper]

if you want to ask something then spit it out...

what's with this close-reopen cycling

[compact pewter]

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[hot moss]

how does one find the answer for bii) I tried and got pi/3 but the answer says it’s 5pi/3

[patent yoke]

Just look at the elements of the matrix and find the theta that works

[hot moss]

doesn’t a=pi/3 work tho?

Cos(pi/3) = 1/2

[patent yoke]

You haven't shown enough work to know what you are doing

[patent yoke]

You can see that B is in the third quadrant

[hot moss]

wait no

[patent yoke]

So -sqrt(3)/2 , -1/2 is 7pi/6

Then you add pi/2 to get 10pi/6

[hot moss]

its not -1/2 tho

[patent yoke]

I'm talking about column 1 of B

You can use column 1 of AB instead if you want, it's the same thing

[hot moss]

I don’t get it

[patent yoke]

Can you see that B is a rotation matrix

[hot moss]

yes

[patent yoke]

The first column is cos(theta) = -sqrt(3)/2, sin(theta) = -1/2.

Using CAST, cos and sin are both negative in the third quadrant

So look it up on the unit circle and you see that theta = 7pi/6

Then A rotates it a further pi/2 so you get 7pi/6 + pi/2 = 7pi/6 + 3pi/6 = 10pi/6 = 5pi/3

Alternatively, you look at column 1 of the matrix AB, cos(theta) = 1/2, sin(theta) = -sqrt(3)/2, cos is positive and sin is negative in the 4th quadrant, and you again get 5pi/3.

[hot moss]

okay I understand the answer I just got it now

but I didn’t learn about using the cast stuff from where I got my notes for this

could u explain the general like solution? like when ur talking about the columns of the matrices

[patent yoke]

CAST is just expected knowledge by the time you are learning this

[hot moss]

well yeah I know CAST but I didn’t know I had to apply it to this question like that

[patent yoke]

There is nothing special about the first column, it's just the entries of a rotation matrix has in its entries cos, sin and -sin

Column 1 just happens to have cos and sin in it, and adding -sin to your system of equations doesn't give you any extra information

[hot moss]

wait I think I’m getting it now

[patent yoke]

For the matrix AB you are just solving the system of equations:

cos(x) = 1/2
sin(x) = -sqrt(3)/2
cos(x) = 1/2
-sin(x) = sqrt(3)/2

The first two equations are from column 1 and the second equations are from column 2

Hopefully it should be obvious that equation 3 and 4 don't help

[hot moss]

Yeah yeah

so instead of just setting the anticlockwise rotation matrix equal to the matrix AB and getting cos(a) = 1/2 which would imply a=pi/3

I use the CAST thing instead to get it

[patent yoke]

You don't really need that CAST rule, it is just a helpful sanity check

[hot moss]

well in general then

[devout latch]

maybe a trig circle could help you?

[hot moss]

but I think I get it

it’s just using the cast diagram for sin and cos?

[patent yoke]

Your problem, maybe, is that cos(pi/3) and cos(5pi/3) are the same value.

and things like CAST help you set up basic expectations for what you think the answer should look like and helps you reduce the chance of an error, but it isn't strictly necessary

[devout latch]

^

[hot moss]

so what ur saying is I don’t have to use the cast diagram?

I mean now that u ahve explained it it seems pretty simple

use it for sin and cos, the answer that appears in both sin and cos is the overall answer to the question

[patent yoke]

I am saying you don't have to use it, but it is good practice to use what you've already learned before to reduce your mistakes.

Solving math problems isn't about just never making an error, it is also about using what you've already learned so you can identify if you've made a mistake on your own.

[hot moss]

I get that but

[patent yoke]

I don't know what is still unclear, we've went through a few different ways to solve it. Just pick the one that you think is easiest.

[hot moss]

no no it’s not that

looking back on it now

I made the mistake of

sina = sqrt3 / 2

Instead of negative

and that’s what probably led me to make that mistake of pi/3

But nah I get it

u gave me the way to solve it and I understand what to do now

so thank you

and also what about C)

I have no clue

[devout latch]

AB rotates by 5pi/3 radian any vector v from R^2 when you do ABv. what fraction of a full turn 2pi radian is that?

[hot moss]

what 😀

[devout latch]

hum maybe the geometric interpretation of what AB does is not a good rabbit hole for now

[devout latch]

reading what was done before, AB is a rotation matrix with theta = 5pi/3. A with theta pi/2 and B with theta 7pi/6. the theta of AB is 7pi/6 + pi/2 = 5pi/3, the sum of the two

that’s true in general the product of rotation matrix is a rotation matrix with theta the sum of each

Now, if theta is a multiple of 2pi the rotation matrix = I

So the question asks when the smallest multiple of 5pi/3 will be a multiple of 2pi

[hot moss]

it does..?

[devout latch]

because (AB)^n will be a rotation matrix with theta n times 5pi/3

by the argument i just did

[hot moss]

this also was definitely not in my notes for matrices…

starting to lose faith rn

[devout latch]

now you know though, the general reason/intuition why this works is that matrices represent linear transformation (functions) transforming the plane R^2 and multiplying them is like composing functions. that’s kinda what i was getting at but that might be too much now

[hot moss]

https://www.advancedhighermaths.co.uk/wp-content/uploads/2017/11/application/pdf/pdf/Matrices-2.pdf

https://www.advancedhighermaths.co.uk/wp-content/uploads/2017/11/application/pdf/pdf/Matrices-1.pdf

if u wanna have a Quick Look through these…

It’s the matrices notes I used and I cannot find anything like that in there

[devout latch]

https://www.youtube.com/watch?v=kYB8IZa5AuE&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=3 if you want to see this in action

[hot moss]

I’ll make sure to have a look, I also got a guy who does these topics for my curriculum which I will also have to look through but thanks for the help

✌🏼

[devout latch]

you have 2 examples of this interpretation of the matrix product

but yeah i agree your question is tricky without this intuition. you would have to multiply (AB) by itself to see the pattern

perhaps

[compact pewter]

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[pallid crypt]

can someone pls tell me how i wud turn this into partial fractions

[pallid crypt]

acc no i got it its okay

lol

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[pallid crypt]

acc no i dont get it

.reppen

.reopen

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[stark marsh]

is the fraction supposed to be u/(u^2-1)^2 ?

[pallid crypt]

4/ (u^2 - 1)^2

mb its so messy haha

late ash

[stark marsh]

i feel like factoring this into
4/[(u-1)^2 (u+1)^2] would make the partial fractions easier

and I think yk the case where some of the expressions reocurr

[pallid crypt]

i see

ill try it and see

.close

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[compact pewter]

@copper loom Has your question been resolved?

[compact pewter]

@copper loom Has your question been resolved?

[honest viper]

Idk if this is true

But here's my proof

Assume Z(G) is not 1

Then there is some nonidentity element g in Z(G)

The conjugacy class of g is just the set with itself

Because it commutes with every element

But <g> is a subset of the conjugacy class

This implies g has degree 1 and is this equal to 1

A contradiction

This |G| = 1 + 3 + 3 + 3 + 5

Also on second thought

I don't think this problem is possible

Because a group of size 15 is always a cyclic group by the sylow theorems, so it must be abelian

But I think my previous proof is the intended solution

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[wary mulch]

are we allowed to use arctan(90 degrees) or no?

[wary mulch]

so basically can we use it for right angled triangles?

[solemn wasp]

arctan is defined for all real numbers, but only returns a number between -pi/2 and pi/2, exclusively

if we expect outputs to be meaningful, we shouldn't feed degree measures into it; they should be like, lengths of line segments

[wary mulch]

can u dumb it down

[solemn wasp]

but if you say that 1 degree = pi/180, then the answer is yes, arctan(pi/2) is a number. it just doesn't mean anything, probably.(?)

[wary mulch]

I dont understand half of the terms

[solemn wasp]

tan is a function; it takes an angle measurement and outputs a real number, which you can think of as the length of a line segment. kind of.

arctan is the inverse of tan: it takes the length of a line segment (again, kind of) and returns an angle measurement

is 90 degrees the kind-of length of a line segment? sure, technically, if you specify that
90 degrees = 90 * (1 degree) = 90 * (pi/180) = pi/2

[solemn wasp]

you probably meant tan(90 degrees). the answer to that is, no, it's undefined.

you can give it other angle measurements from your (right) triangle, just not 90 degrees

[compact pewter]

@wary mulch Has your question been resolved?

[wary mulch]

ohh

how will it be even undefined?

and also for this question i didnt round it to the nearest minute because the question didnt ask for but in the solution, it rounded up to the nearest minute

so im a bit confused

with that

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[wary mulch]

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[glass turret]

hello

[upbeat gorge]

on a right triangle, what's the "opposite" or "adjacent" side of the 90 degrees? both are adjacent to that angle

algebraically, tan(x) = sin(x)/cos(x). What is cos(x) at 90 degrees? Does this cause problems? (hint: yes)

[wary mulch]

do u meant to replace adjacent with hypnoteuse?

because the opposite of a right angle is hypotenuse

[upbeat gorge]

im referring to "adjacent" and "opposite" sides as you would for SOH CAH TOA, if you learned that

i.e. not the hypotenuse

if you had adjacent/hyp, that's just cos, and if you had opposite/hyp, that's just sin

[wary mulch]

true i understand now becuase if right angle was the subject, there would be only 1 oposite, therefore two adjacents

which cant be possible

[upbeat gorge]

i guess you can think of it that way too, but i hesitate to call the hypotenuse the opposite, cause it usually isn't

[wary mulch]

why isnt it?

[upbeat gorge]

well, because the angle is usually touching the hypotenuse and one other side, so the last side is the opposite, and it can't be the hypotenuse ofc

i think you're right and its ok to say that the opposite happens to be the hypotenuse. we are considering an undefined case, after all

but usually, A, O, and H are three different sides, so that's why i hesitate

i'll say for this particular case

[upbeat gorge]

as for this, maybe they said "round all angles to the nearest minute" at the start? if not then ggs

[wary mulch]

jits not touching the hypotenus

e

im talking about this as the subject

u can see its opposite and also the hypotenuse

[upbeat gorge]

yeah yeah, and i meant that we "normally" talk about a or b, so i wasn't sure if we can just immediately say opposite = hypotenuse

but i think it works for this example, and doesn't cause any problems

[compact pewter]

@wary mulch Has your question been resolved?

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[pale trout]

is the * misplaced in the ∂ψ*/∂t on the far right?

[pale trout]

or am i missing something here

cuz this is physics kinda

nvm

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[unkempt sail]

"If g(f(x)) is injective, so is f(x)."
Can I prove this statement with something other than contradiction?

[calm trout]

recite the definition of injectivity

[unkempt sail]

If for some x1,x2 in domain of h(x), h(x1) = h(x2), then x1 = x2 for injectivity

[calm trout]

now prove directly for f

using that g o f is injective

[unkempt sail]

So g o f is injective.
Any x1, x2 in dom(gof), gof(x1) = gof(x2) implies x1 = x2.
How do I continue from here? Shouldn't there also be some restriction on g

[calm trout]

why?

[unkempt sail]

Then how would I continue here

[calm trout]

f(x) = f(y) => ??? => x = y

fill the ???

[unkempt sail]

Didn't you already do that by typing x = y ?

[calm trout]

what?

[unkempt sail]

Also is this f the same f in question or some general f

[feral falcon]

there is an intemediary step which u have to fill in. we dont know that f is injective. that is what we are trying to show

[calm trout]

same f

[unkempt sail]

Ohhh

Do we do g on both sides

Composition with g

[calm trout]

well without that there's no mention of g

so you'd better use g somehow

[unkempt sail]

If f(x) = f(y)
Then g(f(x)) = g(f(y))
Then x = y

[calm trout]

the end

[unkempt sail]

Interesting

[feral falcon]

[unkempt sail]

Ohhh understood

I had a question regarding this method that couldn't we just do this for any general f
Then i realised gof should also be injective

OK thanks

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[jolly crag]

hi, if im finding the gradient of a tangent parallel to the y axis, will the dy/dx value be undefined or will it = 0?

[jolly crag]

my dy/dx is a fraction, so im not sure whether i should equate the numerator or denominator to zero

thank you!

[spice path]

it's undefined

[jolly crag]

so i should equate the denominator to zero right?

[spice path]

sure why not

[jolly crag]

kay thanks

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[raw portal]

how do i find the area

i don't get it intuitively how the area is being traced

at pi/6 the curve passes through origin again

so the part in the 1st quadrant must be traced when were moving from 0 to pi/6

but after pi/6 since r < 0 we get into the third quadrant

then until pi/2 we are in the 3rd quadrant

after pi/2 we get in 4th quadrant and finally at 5pi/6 we are at origin again

dumb

idk

i don't get how the area is being traced out

can someone help

[subtle helm]

If u find it difficult to find how the thing is traced

U can use desmos

[raw portal]

i could figure how areas are traced before this problem

[raw portal]

when solving those problems should i care about it ? how its traced

or just solve the integral find area and move on

i am not very used to checking answers and stuff am lazy

[subtle helm]

Kinda

U need it to set bounds of integration

[raw portal]

yes

[subtle helm]

U can try graphing the major points

And try to visualize the thing

For example at pi/2

The line should be exactly along the y-axis

And at 3pi/2, the thing should also be along the y-axis

[raw portal]

i kinda understand how its being traced

[subtle helm]

Then what is the issue

[raw portal]

nahh

i don't

[raw portal]

yea we can do those things but they don't always help

for example in this problem its not enough

[subtle helm]

ok

so in this problem

[raw portal]

is this how you want me to use desmos ?

[subtle helm]

u can divide the area into 2 parts

[raw portal]

or theres a better way

[subtle helm]

cuz symmetry

the area of one side = integral from area from 0 to pi/6 (the top) + area from 3pi/2 to 2pi - area from pi/6 to pi/2

[subtle helm]

u can set it up like this

for animation

[raw portal]

how

how to set it up for animation

i didnt get that option

[subtle helm]

make thete run from 0 to a

and make a slider for a

[raw portal]

amazing lol

thank you

[raw portal]

i see

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[compact pewter]

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[upbeat geyser]

How the hell do I explain these kinds of questions

[upbeat geyser]

Last time I tried to explain this it took like 20 min

It's gonna take 1 page for me to write the explanation

Context

It's for 3rd and 4th graders

Math Olympiad

[upbeat geyser]

And I didn't even get it completely right

[kind viper]

hmmm

[upbeat geyser]

I hate these types of questions

Trial and error ahh

[kind viper]

C+G=9 i guess

but that's quite a lot of possibilities

[upbeat geyser]

Yeah I understand the explanation but it's just

[kind viper]

though we do know D is either 1 or 2, and A is either 8 or 9

[upbeat geyser]

Really

Really

Long

I'm not willing to write a page full of text

[kind viper]

actually maybe A=9 is not possible cause B+E is at least 1 and we would have sum 9100 or higher for the first addition

[knotty orchid]

its c

[upbeat geyser]

No I know the explanation

Just

How do I explain it to kids

[knotty orchid]

because a is either 8or 9

[upbeat geyser]

3rd and 4th grade

[knotty orchid]

and d is either 1 or 2

[upbeat geyser]

If I wrote a is either 8 or 9 and see how it affects the addition is gonna atleast take up half or 75% of the page

[kind viper]

nah

actually maybe A=9 is not possible cause B+E is at least 1 and we would have sum 9100 or higher for the first addition

you can find D next

then G then C

[knotty orchid]

yeah ann is right

[upbeat geyser]

Lemme try ig

If I can conclude